Mixing
Does a continuous function preserve Bolzano-Weierstrass property?
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Let $X,Y$ be arbitrary topological spaces with $X$ satisfying the Bolzano-Weierstrass property (i.e. every infinite subset has some accumulation point) and let $f:Xto Y$ be a continuous function, does $f(X)$ satisfy the B-W property?
I tried to prove it does by contradiction as follows:
Let's assume $f(X)$ doesn't have the BWP, so there must be some infinite subset $A$ wich has not any accumulation point, hence it's is closed. Because $f$ is continuous, $f^{-1}(A)$ is also closed and infinite and because $X$ has the BWP there is some accumulation point of $f^{-1}(A)$ which is contained in it, let's call $x_0$ one of such points. Now, $f(x_0)$ is an isolated point of A, so ${f(x_0)}$ is an open neighborhood of $f(x_0)$ (with the relative topology of $f(X)$ as a subspace of $Y$).
This means that $f^{-1}(f(x_0))$ is an open neighborhood of $x_0$ which contains other points different than $x_0$. But this doesn't lead me to a contradiction since in principle $f$ could be locally constant at $x_0$. Is there some reason why this cannot happen? Or maybe continuous function doesn't actually preserve BWP, if that's the case could someone provide a counterexample? Constant functions doesn't seem to work because they have finite image, so I thought maybe a locally constant function with infinite image, but I couldn't build up an explicit example.
EDIT to avoid future missunderstandings: What I called 'accumulation point' above is what usually is refered in english as 'limit point'. Explicitly $a$ is an accumulation (limit) point of a set $A$ of a topological space $X$ if every open set containing $a$ intersected with $A$ contains some point different than $a$
general-topology
asked Jul 20 '17 at 22:21
la flacala flaca
1,584518
$endgroup$
add a comment |
$begingroup$
Let $X,Y$ be arbitrary topological spaces with $X$ satisfying the Bolzano-Weierstrass property (i.e. every infinite subset has some accumulation point) and let $f:Xto Y$ be a continuous function, does $f(X)$ satisfy the B-W property?
I tried to prove it does by contradiction as follows:
Let's assume $f(X)$ doesn't have the BWP, so there must be some infinite subset $A$ wich has not any accumulation point, hence it's is closed. Because $f$ is continuous, $f^{-1}(A)$ is also closed and infinite and because $X$ has the BWP there is some accumulation point of $f^{-1}(A)$ which is contained in it, let's call $x_0$ one of such points. Now, $f(x_0)$ is an isolated point of A, so ${f(x_0)}$ is an open neighborhood of $f(x_0)$ (with the relative topology of $f(X)$ as a subspace of $Y$).
This means that $f^{-1}(f(x_0))$ is an open neighborhood of $x_0$ which contains other points different than $x_0$. But this doesn't lead me to a contradiction since in principle $f$ could be locally constant at $x_0$. Is there some reason why this cannot happen? Or maybe continuous function doesn't actually preserve BWP, if that's the case could someone provide a counterexample? Constant functions doesn't seem to work because they have finite image, so I thought maybe a locally constant function with infinite image, but I couldn't build up an explicit example.
EDIT to avoid future missunderstandings: What I called 'accumulation point' above is what usually is refered in english as 'limit point'. Explicitly $a$ is an accumulation (limit) point of a set $A$ of a topological space $X$ if every open set containing $a$ intersected with $A$ contains some point different than $a$
general-topology
asked Jul 20 '17 at 22:21
la flacala flaca
1,584518
$endgroup$
$begingroup$
This reminds me of something similar: Let $X$ be a $T_1$ space and let $f:Xto Y$ be a continuous surjection. Let $D$ be a closed discrete sub-space of $Y.$ Then $X$ has a closed discrete sub-space $C$ of equal cardinal to $D.$ Proof: For $din D$ choose $g(d)in f^{-1}{d}.$ Show that $C=g(D$) is a closed discrete sub-space of $X.$
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– DanielWainfleet
Jul 21 '17 at 2:56
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What is your definition of accumulation point? It seems like a limit point.
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– Henno Brandsma
Jul 21 '17 at 10:26
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My example below shows exactly this locally constant problem...
$endgroup$
– Henno Brandsma
Jul 21 '17 at 11:13
add a comment |
$begingroup$
Let $X,Y$ be arbitrary topological spaces with $X$ satisfying the Bolzano-Weierstrass property (i.e. every infinite subset has some accumulation point) and let $f:Xto Y$ be a continuous function, does $f(X)$ satisfy the B-W property?
I tried to prove it does by contradiction as follows:
Let's assume $f(X)$ doesn't have the BWP, so there must be some infinite subset $A$ wich has not any accumulation point, hence it's is closed. Because $f$ is continuous, $f^{-1}(A)$ is also closed and infinite and because $X$ has the BWP there is some accumulation point of $f^{-1}(A)$ which is contained in it, let's call $x_0$ one of such points. Now, $f(x_0)$ is an isolated point of A, so ${f(x_0)}$ is an open neighborhood of $f(x_0)$ (with the relative topology of $f(X)$ as a subspace of $Y$).
This means that $f^{-1}(f(x_0))$ is an open neighborhood of $x_0$ which contains other points different than $x_0$. But this doesn't lead me to a contradiction since in principle $f$ could be locally constant at $x_0$. Is there some reason why this cannot happen? Or maybe continuous function doesn't actually preserve BWP, if that's the case could someone provide a counterexample? Constant functions doesn't seem to work because they have finite image, so I thought maybe a locally constant function with infinite image, but I couldn't build up an explicit example.
EDIT to avoid future missunderstandings: What I called 'accumulation point' above is what usually is refered in english as 'limit point'. Explicitly $a$ is an accumulation (limit) point of a set $A$ of a topological space $X$ if every open set containing $a$ intersected with $A$ contains some point different than $a$
general-topology
asked Jul 20 '17 at 22:21
la flacala flaca
1,584518
$endgroup$
Let $X,Y$ be arbitrary topological spaces with $X$ satisfying the Bolzano-Weierstrass property (i.e. every infinite subset has some accumulation point) and let $f:Xto Y$ be a continuous function, does $f(X)$ satisfy the B-W property?
I tried to prove it does by contradiction as follows:
Let's assume $f(X)$ doesn't have the BWP, so there must be some infinite subset $A$ wich has not any accumulation point, hence it's is closed. Because $f$ is continuous, $f^{-1}(A)$ is also closed and infinite and because $X$ has the BWP there is some accumulation point of $f^{-1}(A)$ which is contained in it, let's call $x_0$ one of such points. Now, $f(x_0)$ is an isolated point of A, so ${f(x_0)}$ is an open neighborhood of $f(x_0)$ (with the relative topology of $f(X)$ as a subspace of $Y$).
This means that $f^{-1}(f(x_0))$ is an open neighborhood of $x_0$ which contains other points different than $x_0$. But this doesn't lead me to a contradiction since in principle $f$ could be locally constant at $x_0$. Is there some reason why this cannot happen? Or maybe continuous function doesn't actually preserve BWP, if that's the case could someone provide a counterexample? Constant functions doesn't seem to work because they have finite image, so I thought maybe a locally constant function with infinite image, but I couldn't build up an explicit example.
EDIT to avoid future missunderstandings: What I called 'accumulation point' above is what usually is refered in english as 'limit point'. Explicitly $a$ is an accumulation (limit) point of a set $A$ of a topological space $X$ if every open set containing $a$ intersected with $A$ contains some point different than $a$
general-topology
general-topology
asked Jul 20 '17 at 22:21
la flacala flaca
1,584518
asked Jul 20 '17 at 22:21
la flacala flaca
1,584518
edited Jul 21 '17 at 15:34
la flaca
asked Jul 20 '17 at 22:21
la flacala flaca
1,584518
asked Jul 20 '17 at 22:21
la flacala flaca
1,584518
asked Jul 20 '17 at 22:21
la flacala flaca
1,584518
1,584518
$begingroup$
This reminds me of something similar: Let $X$ be a $T_1$ space and let $f:Xto Y$ be a continuous surjection. Let $D$ be a closed discrete sub-space of $Y.$ Then $X$ has a closed discrete sub-space $C$ of equal cardinal to $D.$ Proof: For $din D$ choose $g(d)in f^{-1}{d}.$ Show that $C=g(D$) is a closed discrete sub-space of $X.$
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– DanielWainfleet
Jul 21 '17 at 2:56
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What is your definition of accumulation point? It seems like a limit point.
$endgroup$
– Henno Brandsma
Jul 21 '17 at 10:26
$begingroup$
My example below shows exactly this locally constant problem...
$endgroup$
– Henno Brandsma
Jul 21 '17 at 11:13
add a comment |
$begingroup$
This reminds me of something similar: Let $X$ be a $T_1$ space and let $f:Xto Y$ be a continuous surjection. Let $D$ be a closed discrete sub-space of $Y.$ Then $X$ has a closed discrete sub-space $C$ of equal cardinal to $D.$ Proof: For $din D$ choose $g(d)in f^{-1}{d}.$ Show that $C=g(D$) is a closed discrete sub-space of $X.$
$endgroup$
– DanielWainfleet
Jul 21 '17 at 2:56
$begingroup$
What is your definition of accumulation point? It seems like a limit point.
$endgroup$
– Henno Brandsma
Jul 21 '17 at 10:26
$begingroup$
My example below shows exactly this locally constant problem...
$endgroup$
– Henno Brandsma
Jul 21 '17 at 11:13
$begingroup$
This reminds me of something similar: Let $X$ be a $T_1$ space and let $f:Xto Y$ be a continuous surjection. Let $D$ be a closed discrete sub-space of $Y.$ Then $X$ has a closed discrete sub-space $C$ of equal cardinal to $D.$ Proof: For $din D$ choose $g(d)in f^{-1}{d}.$ Show that $C=g(D$) is a closed discrete sub-space of $X.$
$endgroup$
– DanielWainfleet
Jul 21 '17 at 2:56
$begingroup$
This reminds me of something similar: Let $X$ be a $T_1$ space and let $f:Xto Y$ be a continuous surjection. Let $D$ be a closed discrete sub-space of $Y.$ Then $X$ has a closed discrete sub-space $C$ of equal cardinal to $D.$ Proof: For $din D$ choose $g(d)in f^{-1}{d}.$ Show that $C=g(D$) is a closed discrete sub-space of $X.$
$endgroup$
– DanielWainfleet
Jul 21 '17 at 2:56
$begingroup$
What is your definition of accumulation point? It seems like a limit point.
$endgroup$
– Henno Brandsma
Jul 21 '17 at 10:26
$begingroup$
What is your definition of accumulation point? It seems like a limit point.
$endgroup$
– Henno Brandsma
Jul 21 '17 at 10:26
$begingroup$
My example below shows exactly this locally constant problem...
$endgroup$
– Henno Brandsma
Jul 21 '17 at 11:13
$begingroup$
My example below shows exactly this locally constant problem...
$endgroup$
– Henno Brandsma
Jul 21 '17 at 11:13
add a comment |
3 Answers
3
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This is actually false if "accumulation point" = "limit point" where $x$ is a limit point of $A$ iff every neighbourhood of $x$ intersects $Asetminus{x}$.
Your B-W property is then called "limit point compact"
A counterexample: Let $Y = mathbb{N}$ as a discrete space and let $X = Y times {0,1}$ where ${0,1}$ has the indiscrete (trivial) topology with no non-trivial open sets. Let $f= pi_Y$ be the projection onto $Y$, which is continuous.
$X$ is limit point compact: if $A subseteq X$, and $(x,i) in A$, then $(x,i')$ is a limit point of $A$, where $i' in {0,1}setminus{i}$, because any product open set that contains $(x,i)$ also contains $(x,i')$ and vice versa. But $Y$ is infinite discrete so not limit point compact.
If $X$ is $T_1$ then (here our example is not even $T_0$) limit point compact is equivalent to "strong limit point compact": every infinite set $A$ has an ($omega$-)accumulation point, i.e. an $x in X$ such that every neighbourhood of $x$ contains infinitely many points from $A$. This property is preserved by continuous functions:
Suppose $X$ is strong limit point compact, and $f:X to Y$ is continuous.
Let $A subseteq f[X]$ be infinite and let ${y_n: n in omega}$ be an enumeration ($y_n neq y_m$ for $n neq m$) of a countably infinite subset of $A$ which means that we have for all $n$ we have $x_n in X$ such that $f(x_n) = y_n$. Then clearly, $n neq m$ implies $x_n neq x_m$ as well, so ${x_n: n in omega}$ is an infinite subset of $X$. So this has a strong limit point $x$. Now let $y = f(x) in f[X]$.
If $O$ is any open set containing $y$ then $f^{-1}[O]$ is open by continuity and contains $x$, so there are infinitely many $n$ with $x_n in f^{-1}[O]$. All those $n$ also obey $y_n = f(x_n) in O$. So $y$ is a strong limit point of $A$, and $f[X]$ is strong limit point compact.
So your original statement is true for $T_1$ spaces $X$. We then get even a (slightly) stronger property than B-W in $Y$ (without assuming anything extra on $Y$), namely strong limit compact.
answered Jul 21 '17 at 10:40
Henno BrandsmaHenno Brandsma
106k347114
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Hello there! Yes, my definition of accumulation point coincides with limit point. Sorry for that, I guess it's something cultural, we don't even use that term for sequences (according to wikipedia accumulation point is used as a synonim of cluster point of a sequence, which we call 'agglomeration point' of a sequence)... I like your counterexample, very simple and easy to follow, but I still don't realize what's wrong then with the proof provided by Hellen, I need to think a little more about it... Btw so it is not common in english to say a set 'has the B-W property' to mean what I ment?
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– la flaca
Jul 21 '17 at 14:05
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@laflaca $f(x)$ could be the same as $f(x_n)$
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– Henno Brandsma
Jul 21 '17 at 14:10
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Sure, I got it! ; )
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– la flaca
Jul 21 '17 at 14:20
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@laflaca for me a cluster point of a sequence $(x_n)$ is an $x$ such that for all neighbourhoods $O$ of $x$ and for all $n$ there is some $m > n$ with $x_m in O$. This need not even be a limit point of ${x_n: n inomega}$, we could have a constant subsequence.
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– Henno Brandsma
Jul 21 '17 at 14:40
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sure, I just meant to say that wikipedia agrees with you on the definition of limit point, and it also agrees on that definition of cluster point you gave but it also use the term 'accumulation point' as a synonym of cluster point (which is not the usage I was giving to it). My comment on Hellen's answer identifying a cluster point of a sequence with a limit point of its image is mistaken, precisesely for that reason. But I couldn't edit it, when I was going to post the edit it was too late...
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– la flaca
Jul 21 '17 at 15:27
add a comment |
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Let $f(x_n)=y_nin f(X)$ be a sequence of different points. Then, $x_n$ are different too. Passing to a subsequence, there is an $x$ such that $x_n$ accumulates in $x$. Let $Vni y=f(x)$ be an open subset of $Y$. Then $U=f^{-1}(V)$ is open in $X$ and contains $x$. Therefore some $x_n$ lies in $U$. Therefore $y_n=f(x_n)$ lies in $V$. This means that $y_n$ accumulates in $y$.
edited Jul 21 '17 at 8:28
Rob Arthan
29.2k42966
answered Jul 20 '17 at 22:30
HellenHellen
1165
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Of course, I forgot the usefulness of the axiom of choice. Thank you!
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– la flaca
Jul 20 '17 at 22:41
1
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There is no use of the axiom of choice in this argument.
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– Rob Arthan
Jul 20 '17 at 22:43
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@RobArthan To get $x_n$.
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– Hellen
Jul 20 '17 at 22:45
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@RobArthan There is in my argument.
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– Hellen
Jul 20 '17 at 22:48
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Why the downvote? The answer given by Hellen fixes what was missing in my reasoning. It's great that you can give a proof without using the AC, but this one is still a good answer to my question
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– la flaca
Jul 20 '17 at 22:53
|
show 6 more comments
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Continuity alone has nothing to do with Bolzano-Weierstrass.
Returning to definitions often helps. You need to show that the range of $f$ is well-ordered, and, in addition, continuous.
answered Jan 4 at 23:36
freehumoristfreehumorist
173112
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3 Answers
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3 Answers
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$begingroup$
This is actually false if "accumulation point" = "limit point" where $x$ is a limit point of $A$ iff every neighbourhood of $x$ intersects $Asetminus{x}$.
Your B-W property is then called "limit point compact"
A counterexample: Let $Y = mathbb{N}$ as a discrete space and let $X = Y times {0,1}$ where ${0,1}$ has the indiscrete (trivial) topology with no non-trivial open sets. Let $f= pi_Y$ be the projection onto $Y$, which is continuous.
$X$ is limit point compact: if $A subseteq X$, and $(x,i) in A$, then $(x,i')$ is a limit point of $A$, where $i' in {0,1}setminus{i}$, because any product open set that contains $(x,i)$ also contains $(x,i')$ and vice versa. But $Y$ is infinite discrete so not limit point compact.
If $X$ is $T_1$ then (here our example is not even $T_0$) limit point compact is equivalent to "strong limit point compact": every infinite set $A$ has an ($omega$-)accumulation point, i.e. an $x in X$ such that every neighbourhood of $x$ contains infinitely many points from $A$. This property is preserved by continuous functions:
Suppose $X$ is strong limit point compact, and $f:X to Y$ is continuous.
Let $A subseteq f[X]$ be infinite and let ${y_n: n in omega}$ be an enumeration ($y_n neq y_m$ for $n neq m$) of a countably infinite subset of $A$ which means that we have for all $n$ we have $x_n in X$ such that $f(x_n) = y_n$. Then clearly, $n neq m$ implies $x_n neq x_m$ as well, so ${x_n: n in omega}$ is an infinite subset of $X$. So this has a strong limit point $x$. Now let $y = f(x) in f[X]$.
If $O$ is any open set containing $y$ then $f^{-1}[O]$ is open by continuity and contains $x$, so there are infinitely many $n$ with $x_n in f^{-1}[O]$. All those $n$ also obey $y_n = f(x_n) in O$. So $y$ is a strong limit point of $A$, and $f[X]$ is strong limit point compact.
So your original statement is true for $T_1$ spaces $X$. We then get even a (slightly) stronger property than B-W in $Y$ (without assuming anything extra on $Y$), namely strong limit compact.
answered Jul 21 '17 at 10:40
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
$begingroup$
Hello there! Yes, my definition of accumulation point coincides with limit point. Sorry for that, I guess it's something cultural, we don't even use that term for sequences (according to wikipedia accumulation point is used as a synonim of cluster point of a sequence, which we call 'agglomeration point' of a sequence)... I like your counterexample, very simple and easy to follow, but I still don't realize what's wrong then with the proof provided by Hellen, I need to think a little more about it... Btw so it is not common in english to say a set 'has the B-W property' to mean what I ment?
$endgroup$
– la flaca
Jul 21 '17 at 14:05
$begingroup$
@laflaca $f(x)$ could be the same as $f(x_n)$
$endgroup$
– Henno Brandsma
Jul 21 '17 at 14:10
$begingroup$
Sure, I got it! ; )
$endgroup$
– la flaca
Jul 21 '17 at 14:20
$begingroup$
@laflaca for me a cluster point of a sequence $(x_n)$ is an $x$ such that for all neighbourhoods $O$ of $x$ and for all $n$ there is some $m > n$ with $x_m in O$. This need not even be a limit point of ${x_n: n inomega}$, we could have a constant subsequence.
$endgroup$
– Henno Brandsma
Jul 21 '17 at 14:40
$begingroup$
sure, I just meant to say that wikipedia agrees with you on the definition of limit point, and it also agrees on that definition of cluster point you gave but it also use the term 'accumulation point' as a synonym of cluster point (which is not the usage I was giving to it). My comment on Hellen's answer identifying a cluster point of a sequence with a limit point of its image is mistaken, precisesely for that reason. But I couldn't edit it, when I was going to post the edit it was too late...
$endgroup$
– la flaca
Jul 21 '17 at 15:27
add a comment |
$begingroup$
This is actually false if "accumulation point" = "limit point" where $x$ is a limit point of $A$ iff every neighbourhood of $x$ intersects $Asetminus{x}$.
Your B-W property is then called "limit point compact"
A counterexample: Let $Y = mathbb{N}$ as a discrete space and let $X = Y times {0,1}$ where ${0,1}$ has the indiscrete (trivial) topology with no non-trivial open sets. Let $f= pi_Y$ be the projection onto $Y$, which is continuous.
$X$ is limit point compact: if $A subseteq X$, and $(x,i) in A$, then $(x,i')$ is a limit point of $A$, where $i' in {0,1}setminus{i}$, because any product open set that contains $(x,i)$ also contains $(x,i')$ and vice versa. But $Y$ is infinite discrete so not limit point compact.
If $X$ is $T_1$ then (here our example is not even $T_0$) limit point compact is equivalent to "strong limit point compact": every infinite set $A$ has an ($omega$-)accumulation point, i.e. an $x in X$ such that every neighbourhood of $x$ contains infinitely many points from $A$. This property is preserved by continuous functions:
Suppose $X$ is strong limit point compact, and $f:X to Y$ is continuous.
Let $A subseteq f[X]$ be infinite and let ${y_n: n in omega}$ be an enumeration ($y_n neq y_m$ for $n neq m$) of a countably infinite subset of $A$ which means that we have for all $n$ we have $x_n in X$ such that $f(x_n) = y_n$. Then clearly, $n neq m$ implies $x_n neq x_m$ as well, so ${x_n: n in omega}$ is an infinite subset of $X$. So this has a strong limit point $x$. Now let $y = f(x) in f[X]$.
If $O$ is any open set containing $y$ then $f^{-1}[O]$ is open by continuity and contains $x$, so there are infinitely many $n$ with $x_n in f^{-1}[O]$. All those $n$ also obey $y_n = f(x_n) in O$. So $y$ is a strong limit point of $A$, and $f[X]$ is strong limit point compact.
So your original statement is true for $T_1$ spaces $X$. We then get even a (slightly) stronger property than B-W in $Y$ (without assuming anything extra on $Y$), namely strong limit compact.
answered Jul 21 '17 at 10:40
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
$begingroup$
Hello there! Yes, my definition of accumulation point coincides with limit point. Sorry for that, I guess it's something cultural, we don't even use that term for sequences (according to wikipedia accumulation point is used as a synonim of cluster point of a sequence, which we call 'agglomeration point' of a sequence)... I like your counterexample, very simple and easy to follow, but I still don't realize what's wrong then with the proof provided by Hellen, I need to think a little more about it... Btw so it is not common in english to say a set 'has the B-W property' to mean what I ment?
$endgroup$
– la flaca
Jul 21 '17 at 14:05
$begingroup$
@laflaca $f(x)$ could be the same as $f(x_n)$
$endgroup$
– Henno Brandsma
Jul 21 '17 at 14:10
$begingroup$
Sure, I got it! ; )
$endgroup$
– la flaca
Jul 21 '17 at 14:20
$begingroup$
@laflaca for me a cluster point of a sequence $(x_n)$ is an $x$ such that for all neighbourhoods $O$ of $x$ and for all $n$ there is some $m > n$ with $x_m in O$. This need not even be a limit point of ${x_n: n inomega}$, we could have a constant subsequence.
$endgroup$
– Henno Brandsma
Jul 21 '17 at 14:40
$begingroup$
sure, I just meant to say that wikipedia agrees with you on the definition of limit point, and it also agrees on that definition of cluster point you gave but it also use the term 'accumulation point' as a synonym of cluster point (which is not the usage I was giving to it). My comment on Hellen's answer identifying a cluster point of a sequence with a limit point of its image is mistaken, precisesely for that reason. But I couldn't edit it, when I was going to post the edit it was too late...
$endgroup$
– la flaca
Jul 21 '17 at 15:27
add a comment |
$begingroup$
This is actually false if "accumulation point" = "limit point" where $x$ is a limit point of $A$ iff every neighbourhood of $x$ intersects $Asetminus{x}$.
Your B-W property is then called "limit point compact"
A counterexample: Let $Y = mathbb{N}$ as a discrete space and let $X = Y times {0,1}$ where ${0,1}$ has the indiscrete (trivial) topology with no non-trivial open sets. Let $f= pi_Y$ be the projection onto $Y$, which is continuous.
$X$ is limit point compact: if $A subseteq X$, and $(x,i) in A$, then $(x,i')$ is a limit point of $A$, where $i' in {0,1}setminus{i}$, because any product open set that contains $(x,i)$ also contains $(x,i')$ and vice versa. But $Y$ is infinite discrete so not limit point compact.
If $X$ is $T_1$ then (here our example is not even $T_0$) limit point compact is equivalent to "strong limit point compact": every infinite set $A$ has an ($omega$-)accumulation point, i.e. an $x in X$ such that every neighbourhood of $x$ contains infinitely many points from $A$. This property is preserved by continuous functions:
Suppose $X$ is strong limit point compact, and $f:X to Y$ is continuous.
Let $A subseteq f[X]$ be infinite and let ${y_n: n in omega}$ be an enumeration ($y_n neq y_m$ for $n neq m$) of a countably infinite subset of $A$ which means that we have for all $n$ we have $x_n in X$ such that $f(x_n) = y_n$. Then clearly, $n neq m$ implies $x_n neq x_m$ as well, so ${x_n: n in omega}$ is an infinite subset of $X$. So this has a strong limit point $x$. Now let $y = f(x) in f[X]$.
If $O$ is any open set containing $y$ then $f^{-1}[O]$ is open by continuity and contains $x$, so there are infinitely many $n$ with $x_n in f^{-1}[O]$. All those $n$ also obey $y_n = f(x_n) in O$. So $y$ is a strong limit point of $A$, and $f[X]$ is strong limit point compact.
So your original statement is true for $T_1$ spaces $X$. We then get even a (slightly) stronger property than B-W in $Y$ (without assuming anything extra on $Y$), namely strong limit compact.
answered Jul 21 '17 at 10:40
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
This is actually false if "accumulation point" = "limit point" where $x$ is a limit point of $A$ iff every neighbourhood of $x$ intersects $Asetminus{x}$.
Your B-W property is then called "limit point compact"
A counterexample: Let $Y = mathbb{N}$ as a discrete space and let $X = Y times {0,1}$ where ${0,1}$ has the indiscrete (trivial) topology with no non-trivial open sets. Let $f= pi_Y$ be the projection onto $Y$, which is continuous.
$X$ is limit point compact: if $A subseteq X$, and $(x,i) in A$, then $(x,i')$ is a limit point of $A$, where $i' in {0,1}setminus{i}$, because any product open set that contains $(x,i)$ also contains $(x,i')$ and vice versa. But $Y$ is infinite discrete so not limit point compact.
If $X$ is $T_1$ then (here our example is not even $T_0$) limit point compact is equivalent to "strong limit point compact": every infinite set $A$ has an ($omega$-)accumulation point, i.e. an $x in X$ such that every neighbourhood of $x$ contains infinitely many points from $A$. This property is preserved by continuous functions:
Suppose $X$ is strong limit point compact, and $f:X to Y$ is continuous.
Let $A subseteq f[X]$ be infinite and let ${y_n: n in omega}$ be an enumeration ($y_n neq y_m$ for $n neq m$) of a countably infinite subset of $A$ which means that we have for all $n$ we have $x_n in X$ such that $f(x_n) = y_n$. Then clearly, $n neq m$ implies $x_n neq x_m$ as well, so ${x_n: n in omega}$ is an infinite subset of $X$. So this has a strong limit point $x$. Now let $y = f(x) in f[X]$.
If $O$ is any open set containing $y$ then $f^{-1}[O]$ is open by continuity and contains $x$, so there are infinitely many $n$ with $x_n in f^{-1}[O]$. All those $n$ also obey $y_n = f(x_n) in O$. So $y$ is a strong limit point of $A$, and $f[X]$ is strong limit point compact.
So your original statement is true for $T_1$ spaces $X$. We then get even a (slightly) stronger property than B-W in $Y$ (without assuming anything extra on $Y$), namely strong limit compact.
answered Jul 21 '17 at 10:40
Henno BrandsmaHenno Brandsma
106k347114
edited Jul 21 '17 at 11:11
answered Jul 21 '17 at 10:40
Henno BrandsmaHenno Brandsma
106k347114
answered Jul 21 '17 at 10:40
Henno BrandsmaHenno Brandsma
106k347114
answered Jul 21 '17 at 10:40
Henno BrandsmaHenno Brandsma
106k347114
106k347114
$begingroup$
Hello there! Yes, my definition of accumulation point coincides with limit point. Sorry for that, I guess it's something cultural, we don't even use that term for sequences (according to wikipedia accumulation point is used as a synonim of cluster point of a sequence, which we call 'agglomeration point' of a sequence)... I like your counterexample, very simple and easy to follow, but I still don't realize what's wrong then with the proof provided by Hellen, I need to think a little more about it... Btw so it is not common in english to say a set 'has the B-W property' to mean what I ment?
$endgroup$
– la flaca
Jul 21 '17 at 14:05
$begingroup$
@laflaca $f(x)$ could be the same as $f(x_n)$
$endgroup$
– Henno Brandsma
Jul 21 '17 at 14:10
$begingroup$
Sure, I got it! ; )
$endgroup$
– la flaca
Jul 21 '17 at 14:20
$begingroup$
@laflaca for me a cluster point of a sequence $(x_n)$ is an $x$ such that for all neighbourhoods $O$ of $x$ and for all $n$ there is some $m > n$ with $x_m in O$. This need not even be a limit point of ${x_n: n inomega}$, we could have a constant subsequence.
$endgroup$
– Henno Brandsma
Jul 21 '17 at 14:40
$begingroup$
sure, I just meant to say that wikipedia agrees with you on the definition of limit point, and it also agrees on that definition of cluster point you gave but it also use the term 'accumulation point' as a synonym of cluster point (which is not the usage I was giving to it). My comment on Hellen's answer identifying a cluster point of a sequence with a limit point of its image is mistaken, precisesely for that reason. But I couldn't edit it, when I was going to post the edit it was too late...
$endgroup$
– la flaca
Jul 21 '17 at 15:27
add a comment |
$begingroup$
Hello there! Yes, my definition of accumulation point coincides with limit point. Sorry for that, I guess it's something cultural, we don't even use that term for sequences (according to wikipedia accumulation point is used as a synonim of cluster point of a sequence, which we call 'agglomeration point' of a sequence)... I like your counterexample, very simple and easy to follow, but I still don't realize what's wrong then with the proof provided by Hellen, I need to think a little more about it... Btw so it is not common in english to say a set 'has the B-W property' to mean what I ment?
$endgroup$
– la flaca
Jul 21 '17 at 14:05
$begingroup$
@laflaca $f(x)$ could be the same as $f(x_n)$
$endgroup$
– Henno Brandsma
Jul 21 '17 at 14:10
$begingroup$
Sure, I got it! ; )
$endgroup$
– la flaca
Jul 21 '17 at 14:20
$begingroup$
@laflaca for me a cluster point of a sequence $(x_n)$ is an $x$ such that for all neighbourhoods $O$ of $x$ and for all $n$ there is some $m > n$ with $x_m in O$. This need not even be a limit point of ${x_n: n inomega}$, we could have a constant subsequence.
$endgroup$
– Henno Brandsma
Jul 21 '17 at 14:40
$begingroup$
sure, I just meant to say that wikipedia agrees with you on the definition of limit point, and it also agrees on that definition of cluster point you gave but it also use the term 'accumulation point' as a synonym of cluster point (which is not the usage I was giving to it). My comment on Hellen's answer identifying a cluster point of a sequence with a limit point of its image is mistaken, precisesely for that reason. But I couldn't edit it, when I was going to post the edit it was too late...
$endgroup$
– la flaca
Jul 21 '17 at 15:27
$begingroup$
Hello there! Yes, my definition of accumulation point coincides with limit point. Sorry for that, I guess it's something cultural, we don't even use that term for sequences (according to wikipedia accumulation point is used as a synonim of cluster point of a sequence, which we call 'agglomeration point' of a sequence)... I like your counterexample, very simple and easy to follow, but I still don't realize what's wrong then with the proof provided by Hellen, I need to think a little more about it... Btw so it is not common in english to say a set 'has the B-W property' to mean what I ment?
$endgroup$
– la flaca
Jul 21 '17 at 14:05
$begingroup$
Hello there! Yes, my definition of accumulation point coincides with limit point. Sorry for that, I guess it's something cultural, we don't even use that term for sequences (according to wikipedia accumulation point is used as a synonim of cluster point of a sequence, which we call 'agglomeration point' of a sequence)... I like your counterexample, very simple and easy to follow, but I still don't realize what's wrong then with the proof provided by Hellen, I need to think a little more about it... Btw so it is not common in english to say a set 'has the B-W property' to mean what I ment?
$endgroup$
– la flaca
Jul 21 '17 at 14:05
$begingroup$
@laflaca $f(x)$ could be the same as $f(x_n)$
$endgroup$
– Henno Brandsma
Jul 21 '17 at 14:10
$begingroup$
@laflaca $f(x)$ could be the same as $f(x_n)$
$endgroup$
– Henno Brandsma
Jul 21 '17 at 14:10
$begingroup$
Sure, I got it! ; )
$endgroup$
– la flaca
Jul 21 '17 at 14:20
$begingroup$
Sure, I got it! ; )
$endgroup$
– la flaca
Jul 21 '17 at 14:20
$begingroup$
@laflaca for me a cluster point of a sequence $(x_n)$ is an $x$ such that for all neighbourhoods $O$ of $x$ and for all $n$ there is some $m > n$ with $x_m in O$. This need not even be a limit point of ${x_n: n inomega}$, we could have a constant subsequence.
$endgroup$
– Henno Brandsma
Jul 21 '17 at 14:40
$begingroup$
@laflaca for me a cluster point of a sequence $(x_n)$ is an $x$ such that for all neighbourhoods $O$ of $x$ and for all $n$ there is some $m > n$ with $x_m in O$. This need not even be a limit point of ${x_n: n inomega}$, we could have a constant subsequence.
$endgroup$
– Henno Brandsma
Jul 21 '17 at 14:40
$begingroup$
sure, I just meant to say that wikipedia agrees with you on the definition of limit point, and it also agrees on that definition of cluster point you gave but it also use the term 'accumulation point' as a synonym of cluster point (which is not the usage I was giving to it). My comment on Hellen's answer identifying a cluster point of a sequence with a limit point of its image is mistaken, precisesely for that reason. But I couldn't edit it, when I was going to post the edit it was too late...
$endgroup$
– la flaca
Jul 21 '17 at 15:27
$begingroup$
sure, I just meant to say that wikipedia agrees with you on the definition of limit point, and it also agrees on that definition of cluster point you gave but it also use the term 'accumulation point' as a synonym of cluster point (which is not the usage I was giving to it). My comment on Hellen's answer identifying a cluster point of a sequence with a limit point of its image is mistaken, precisesely for that reason. But I couldn't edit it, when I was going to post the edit it was too late...
$endgroup$
– la flaca
Jul 21 '17 at 15:27
add a comment |
$begingroup$
Let $f(x_n)=y_nin f(X)$ be a sequence of different points. Then, $x_n$ are different too. Passing to a subsequence, there is an $x$ such that $x_n$ accumulates in $x$. Let $Vni y=f(x)$ be an open subset of $Y$. Then $U=f^{-1}(V)$ is open in $X$ and contains $x$. Therefore some $x_n$ lies in $U$. Therefore $y_n=f(x_n)$ lies in $V$. This means that $y_n$ accumulates in $y$.
edited Jul 21 '17 at 8:28
Rob Arthan
29.2k42966
answered Jul 20 '17 at 22:30
HellenHellen
1165
$endgroup$
$begingroup$
Of course, I forgot the usefulness of the axiom of choice. Thank you!
$endgroup$
– la flaca
Jul 20 '17 at 22:41
1
$begingroup$
There is no use of the axiom of choice in this argument.
$endgroup$
– Rob Arthan
Jul 20 '17 at 22:43
$begingroup$
@RobArthan To get $x_n$.
$endgroup$
– Hellen
Jul 20 '17 at 22:45
$begingroup$
@RobArthan There is in my argument.
$endgroup$
– Hellen
Jul 20 '17 at 22:48
$begingroup$
Why the downvote? The answer given by Hellen fixes what was missing in my reasoning. It's great that you can give a proof without using the AC, but this one is still a good answer to my question
$endgroup$
– la flaca
Jul 20 '17 at 22:53
|
show 6 more comments
$begingroup$
Let $f(x_n)=y_nin f(X)$ be a sequence of different points. Then, $x_n$ are different too. Passing to a subsequence, there is an $x$ such that $x_n$ accumulates in $x$. Let $Vni y=f(x)$ be an open subset of $Y$. Then $U=f^{-1}(V)$ is open in $X$ and contains $x$. Therefore some $x_n$ lies in $U$. Therefore $y_n=f(x_n)$ lies in $V$. This means that $y_n$ accumulates in $y$.
edited Jul 21 '17 at 8:28
Rob Arthan
29.2k42966
answered Jul 20 '17 at 22:30
HellenHellen
1165
$endgroup$
$begingroup$
Of course, I forgot the usefulness of the axiom of choice. Thank you!
$endgroup$
– la flaca
Jul 20 '17 at 22:41
1
$begingroup$
There is no use of the axiom of choice in this argument.
$endgroup$
– Rob Arthan
Jul 20 '17 at 22:43
$begingroup$
@RobArthan To get $x_n$.
$endgroup$
– Hellen
Jul 20 '17 at 22:45
$begingroup$
@RobArthan There is in my argument.
$endgroup$
– Hellen
Jul 20 '17 at 22:48
$begingroup$
Why the downvote? The answer given by Hellen fixes what was missing in my reasoning. It's great that you can give a proof without using the AC, but this one is still a good answer to my question
$endgroup$
– la flaca
Jul 20 '17 at 22:53
|
show 6 more comments
$begingroup$
Let $f(x_n)=y_nin f(X)$ be a sequence of different points. Then, $x_n$ are different too. Passing to a subsequence, there is an $x$ such that $x_n$ accumulates in $x$. Let $Vni y=f(x)$ be an open subset of $Y$. Then $U=f^{-1}(V)$ is open in $X$ and contains $x$. Therefore some $x_n$ lies in $U$. Therefore $y_n=f(x_n)$ lies in $V$. This means that $y_n$ accumulates in $y$.
edited Jul 21 '17 at 8:28
Rob Arthan
29.2k42966
answered Jul 20 '17 at 22:30
HellenHellen
1165
$endgroup$
Let $f(x_n)=y_nin f(X)$ be a sequence of different points. Then, $x_n$ are different too. Passing to a subsequence, there is an $x$ such that $x_n$ accumulates in $x$. Let $Vni y=f(x)$ be an open subset of $Y$. Then $U=f^{-1}(V)$ is open in $X$ and contains $x$. Therefore some $x_n$ lies in $U$. Therefore $y_n=f(x_n)$ lies in $V$. This means that $y_n$ accumulates in $y$.
edited Jul 21 '17 at 8:28
Rob Arthan
29.2k42966
answered Jul 20 '17 at 22:30
HellenHellen
1165
edited Jul 21 '17 at 8:28
Rob Arthan
29.2k42966
edited Jul 21 '17 at 8:28
Rob Arthan
29.2k42966
edited Jul 21 '17 at 8:28
Rob Arthan
29.2k42966
29.2k42966
answered Jul 20 '17 at 22:30
HellenHellen
1165
answered Jul 20 '17 at 22:30
HellenHellen
1165
answered Jul 20 '17 at 22:30
HellenHellen
1165
1165
$begingroup$
Of course, I forgot the usefulness of the axiom of choice. Thank you!
$endgroup$
– la flaca
Jul 20 '17 at 22:41
1
$begingroup$
There is no use of the axiom of choice in this argument.
$endgroup$
– Rob Arthan
Jul 20 '17 at 22:43
$begingroup$
@RobArthan To get $x_n$.
$endgroup$
– Hellen
Jul 20 '17 at 22:45
$begingroup$
@RobArthan There is in my argument.
$endgroup$
– Hellen
Jul 20 '17 at 22:48
$begingroup$
Why the downvote? The answer given by Hellen fixes what was missing in my reasoning. It's great that you can give a proof without using the AC, but this one is still a good answer to my question
$endgroup$
– la flaca
Jul 20 '17 at 22:53
|
show 6 more comments
$begingroup$
Of course, I forgot the usefulness of the axiom of choice. Thank you!
$endgroup$
– la flaca
Jul 20 '17 at 22:41
1
$begingroup$
There is no use of the axiom of choice in this argument.
$endgroup$
– Rob Arthan
Jul 20 '17 at 22:43
$begingroup$
@RobArthan To get $x_n$.
$endgroup$
– Hellen
Jul 20 '17 at 22:45
$begingroup$
@RobArthan There is in my argument.
$endgroup$
– Hellen
Jul 20 '17 at 22:48
$begingroup$
Why the downvote? The answer given by Hellen fixes what was missing in my reasoning. It's great that you can give a proof without using the AC, but this one is still a good answer to my question
$endgroup$
– la flaca
Jul 20 '17 at 22:53
$begingroup$
Of course, I forgot the usefulness of the axiom of choice. Thank you!
$endgroup$
– la flaca
Jul 20 '17 at 22:41
$begingroup$
Of course, I forgot the usefulness of the axiom of choice. Thank you!
$endgroup$
– la flaca
Jul 20 '17 at 22:41
1
1
$begingroup$
There is no use of the axiom of choice in this argument.
$endgroup$
– Rob Arthan
Jul 20 '17 at 22:43
$begingroup$
There is no use of the axiom of choice in this argument.
$endgroup$
– Rob Arthan
Jul 20 '17 at 22:43
$begingroup$
@RobArthan To get $x_n$.
$endgroup$
– Hellen
Jul 20 '17 at 22:45
$begingroup$
@RobArthan To get $x_n$.
$endgroup$
– Hellen
Jul 20 '17 at 22:45
$begingroup$
@RobArthan There is in my argument.
$endgroup$
– Hellen
Jul 20 '17 at 22:48
$begingroup$
@RobArthan There is in my argument.
$endgroup$
– Hellen
Jul 20 '17 at 22:48
$begingroup$
Why the downvote? The answer given by Hellen fixes what was missing in my reasoning. It's great that you can give a proof without using the AC, but this one is still a good answer to my question
$endgroup$
– la flaca
Jul 20 '17 at 22:53
$begingroup$
Why the downvote? The answer given by Hellen fixes what was missing in my reasoning. It's great that you can give a proof without using the AC, but this one is still a good answer to my question
$endgroup$
– la flaca
Jul 20 '17 at 22:53
|
show 6 more comments
$begingroup$
Continuity alone has nothing to do with Bolzano-Weierstrass.
Returning to definitions often helps. You need to show that the range of $f$ is well-ordered, and, in addition, continuous.
answered Jan 4 at 23:36
freehumoristfreehumorist
173112
$endgroup$
add a comment |
$begingroup$
Continuity alone has nothing to do with Bolzano-Weierstrass.
Returning to definitions often helps. You need to show that the range of $f$ is well-ordered, and, in addition, continuous.
answered Jan 4 at 23:36
freehumoristfreehumorist
173112
$endgroup$
add a comment |
$begingroup$
Continuity alone has nothing to do with Bolzano-Weierstrass.
Returning to definitions often helps. You need to show that the range of $f$ is well-ordered, and, in addition, continuous.
answered Jan 4 at 23:36
freehumoristfreehumorist
173112
$endgroup$
Continuity alone has nothing to do with Bolzano-Weierstrass.
Returning to definitions often helps. You need to show that the range of $f$ is well-ordered, and, in addition, continuous.
answered Jan 4 at 23:36
freehumoristfreehumorist
173112
answered Jan 4 at 23:36
freehumoristfreehumorist
173112
answered Jan 4 at 23:36
freehumoristfreehumorist
173112
answered Jan 4 at 23:36
freehumoristfreehumorist
173112
173112
add a comment |
add a comment |
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$begingroup$
This reminds me of something similar: Let $X$ be a $T_1$ space and let $f:Xto Y$ be a continuous surjection. Let $D$ be a closed discrete sub-space of $Y.$ Then $X$ has a closed discrete sub-space $C$ of equal cardinal to $D.$ Proof: For $din D$ choose $g(d)in f^{-1}{d}.$ Show that $C=g(D$) is a closed discrete sub-space of $X.$
$endgroup$
– DanielWainfleet
Jul 21 '17 at 2:56
$begingroup$
What is your definition of accumulation point? It seems like a limit point.
$endgroup$
– Henno Brandsma
Jul 21 '17 at 10:26
$begingroup$
My example below shows exactly this locally constant problem...
$endgroup$
– Henno Brandsma
Jul 21 '17 at 11:13