Mixing
Prove $f(overline{U})) subset overline{f(U)}$
$begingroup$
I'm facing this problem: For $f : (X,mathcal{T}) to (Y,mathcal{T}')$ a continuous map between topological space, then for any subset $U subset X$ is $f(overline{U)) subset overline{f(U)}$.
I know the usual proof, but I´d like is this one is right:
Let be $x in f(overline{U})$. So for any open neighborhood $V$ of $x$ is $V cap f(U) neq emptyset$.
Since $f$ is continuous, $f^{-1}(V)$ is an open neighborhood of $f^{-1}(x)$.
Using $f^{-1}(x) in overline{U}$ is $f^{-1}(V) cap U neq emptyset$.
So $f(f^{-1}(V) cap U) = V cap f(U) neq emptyset$, then $x in overline{f(U)}$.
Is this right? I'm not sure.
And here another try:
Suppose exist $x in f(overline{U})$ with $x notin overline{f(U)}.$ Then exist an open neightborhood $V$ of $x$ such that $V cap f(U) = emptyset$.
For $z in overline{U}$ such that $f(z) = x$, for any $W$ open neighborhood of $z$ is $W cap A neq emptyset$.
Then exist $h in W cap A$ and $f(h) in f(W) cap f(A)$. So for the adequate $f(W) subset V$ is $f(W) cap U neq emptyset$ but $V cap f(U) = emptyset$. A contradiction.
Is this last one right?
Thanks!
general-topology
asked Jan 22 at 19:11
LH8LH8
1438
$endgroup$
add a comment |
$begingroup$
I'm facing this problem: For $f : (X,mathcal{T}) to (Y,mathcal{T}')$ a continuous map between topological space, then for any subset $U subset X$ is $f(overline{U)) subset overline{f(U)}$.
I know the usual proof, but I´d like is this one is right:
Let be $x in f(overline{U})$. So for any open neighborhood $V$ of $x$ is $V cap f(U) neq emptyset$.
Since $f$ is continuous, $f^{-1}(V)$ is an open neighborhood of $f^{-1}(x)$.
Using $f^{-1}(x) in overline{U}$ is $f^{-1}(V) cap U neq emptyset$.
So $f(f^{-1}(V) cap U) = V cap f(U) neq emptyset$, then $x in overline{f(U)}$.
Is this right? I'm not sure.
And here another try:
Suppose exist $x in f(overline{U})$ with $x notin overline{f(U)}.$ Then exist an open neightborhood $V$ of $x$ such that $V cap f(U) = emptyset$.
For $z in overline{U}$ such that $f(z) = x$, for any $W$ open neighborhood of $z$ is $W cap A neq emptyset$.
Then exist $h in W cap A$ and $f(h) in f(W) cap f(A)$. So for the adequate $f(W) subset V$ is $f(W) cap U neq emptyset$ but $V cap f(U) = emptyset$. A contradiction.
Is this last one right?
Thanks!
general-topology
asked Jan 22 at 19:11
LH8LH8
1438
$endgroup$
1
$begingroup$
Be careful about which space your neighborhoods are in. You write "let $xin f(overline U)$. So for any open neigborhood $V$ of $x$ is $Vcap f(U)neq varnothing$". This is not right. What you really have is that $x = f(x')$ for some $xin X$ such that for every open neigborhood $Vni x$, $Vcap Uneqvarnothing$. EDIT: Sorry, when you said "so [etc]", I thought you claimed that [etc] was a consequence.
$endgroup$
– o.h.
Jan 22 at 19:25
$begingroup$
In general is true $f(U cap V) = f(U) cap f(V)$?
$endgroup$
– LH8
Jan 22 at 20:02
$begingroup$
If $f$ is not injective, $U cap V$ could be empty even if $f(U) cap f(V)$ is not.
$endgroup$
– Umberto P.
Jan 22 at 20:05
$begingroup$
And the converse? $f(U) cap f(V)$ not empty but $U cap V$ yes?
$endgroup$
– LH8
Jan 22 at 20:19
add a comment |
$begingroup$
I'm facing this problem: For $f : (X,mathcal{T}) to (Y,mathcal{T}')$ a continuous map between topological space, then for any subset $U subset X$ is $f(overline{U)) subset overline{f(U)}$.
I know the usual proof, but I´d like is this one is right:
Let be $x in f(overline{U})$. So for any open neighborhood $V$ of $x$ is $V cap f(U) neq emptyset$.
Since $f$ is continuous, $f^{-1}(V)$ is an open neighborhood of $f^{-1}(x)$.
Using $f^{-1}(x) in overline{U}$ is $f^{-1}(V) cap U neq emptyset$.
So $f(f^{-1}(V) cap U) = V cap f(U) neq emptyset$, then $x in overline{f(U)}$.
Is this right? I'm not sure.
And here another try:
Suppose exist $x in f(overline{U})$ with $x notin overline{f(U)}.$ Then exist an open neightborhood $V$ of $x$ such that $V cap f(U) = emptyset$.
For $z in overline{U}$ such that $f(z) = x$, for any $W$ open neighborhood of $z$ is $W cap A neq emptyset$.
Then exist $h in W cap A$ and $f(h) in f(W) cap f(A)$. So for the adequate $f(W) subset V$ is $f(W) cap U neq emptyset$ but $V cap f(U) = emptyset$. A contradiction.
Is this last one right?
Thanks!
general-topology
asked Jan 22 at 19:11
LH8LH8
1438
$endgroup$
I'm facing this problem: For $f : (X,mathcal{T}) to (Y,mathcal{T}')$ a continuous map between topological space, then for any subset $U subset X$ is $f(overline{U)) subset overline{f(U)}$.
I know the usual proof, but I´d like is this one is right:
Let be $x in f(overline{U})$. So for any open neighborhood $V$ of $x$ is $V cap f(U) neq emptyset$.
Since $f$ is continuous, $f^{-1}(V)$ is an open neighborhood of $f^{-1}(x)$.
Using $f^{-1}(x) in overline{U}$ is $f^{-1}(V) cap U neq emptyset$.
So $f(f^{-1}(V) cap U) = V cap f(U) neq emptyset$, then $x in overline{f(U)}$.
Is this right? I'm not sure.
And here another try:
Suppose exist $x in f(overline{U})$ with $x notin overline{f(U)}.$ Then exist an open neightborhood $V$ of $x$ such that $V cap f(U) = emptyset$.
For $z in overline{U}$ such that $f(z) = x$, for any $W$ open neighborhood of $z$ is $W cap A neq emptyset$.
Then exist $h in W cap A$ and $f(h) in f(W) cap f(A)$. So for the adequate $f(W) subset V$ is $f(W) cap U neq emptyset$ but $V cap f(U) = emptyset$. A contradiction.
Is this last one right?
Thanks!
general-topology
general-topology
asked Jan 22 at 19:11
LH8LH8
1438
asked Jan 22 at 19:11
LH8LH8
1438
edited Jan 22 at 19:20
LH8
asked Jan 22 at 19:11
LH8LH8
1438
asked Jan 22 at 19:11
LH8LH8
1438
asked Jan 22 at 19:11
LH8LH8
1438
1438
1
$begingroup$
Be careful about which space your neighborhoods are in. You write "let $xin f(overline U)$. So for any open neigborhood $V$ of $x$ is $Vcap f(U)neq varnothing$". This is not right. What you really have is that $x = f(x')$ for some $xin X$ such that for every open neigborhood $Vni x$, $Vcap Uneqvarnothing$. EDIT: Sorry, when you said "so [etc]", I thought you claimed that [etc] was a consequence.
$endgroup$
– o.h.
Jan 22 at 19:25
$begingroup$
In general is true $f(U cap V) = f(U) cap f(V)$?
$endgroup$
– LH8
Jan 22 at 20:02
$begingroup$
If $f$ is not injective, $U cap V$ could be empty even if $f(U) cap f(V)$ is not.
$endgroup$
– Umberto P.
Jan 22 at 20:05
$begingroup$
And the converse? $f(U) cap f(V)$ not empty but $U cap V$ yes?
$endgroup$
– LH8
Jan 22 at 20:19
add a comment |
1
$begingroup$
Be careful about which space your neighborhoods are in. You write "let $xin f(overline U)$. So for any open neigborhood $V$ of $x$ is $Vcap f(U)neq varnothing$". This is not right. What you really have is that $x = f(x')$ for some $xin X$ such that for every open neigborhood $Vni x$, $Vcap Uneqvarnothing$. EDIT: Sorry, when you said "so [etc]", I thought you claimed that [etc] was a consequence.
$endgroup$
– o.h.
Jan 22 at 19:25
$begingroup$
In general is true $f(U cap V) = f(U) cap f(V)$?
$endgroup$
– LH8
Jan 22 at 20:02
$begingroup$
If $f$ is not injective, $U cap V$ could be empty even if $f(U) cap f(V)$ is not.
$endgroup$
– Umberto P.
Jan 22 at 20:05
$begingroup$
And the converse? $f(U) cap f(V)$ not empty but $U cap V$ yes?
$endgroup$
– LH8
Jan 22 at 20:19
1
1
$begingroup$
Be careful about which space your neighborhoods are in. You write "let $xin f(overline U)$. So for any open neigborhood $V$ of $x$ is $Vcap f(U)neq varnothing$". This is not right. What you really have is that $x = f(x')$ for some $xin X$ such that for every open neigborhood $Vni x$, $Vcap Uneqvarnothing$. EDIT: Sorry, when you said "so [etc]", I thought you claimed that [etc] was a consequence.
$endgroup$
– o.h.
Jan 22 at 19:25
$begingroup$
Be careful about which space your neighborhoods are in. You write "let $xin f(overline U)$. So for any open neigborhood $V$ of $x$ is $Vcap f(U)neq varnothing$". This is not right. What you really have is that $x = f(x')$ for some $xin X$ such that for every open neigborhood $Vni x$, $Vcap Uneqvarnothing$. EDIT: Sorry, when you said "so [etc]", I thought you claimed that [etc] was a consequence.
$endgroup$
– o.h.
Jan 22 at 19:25
$begingroup$
In general is true $f(U cap V) = f(U) cap f(V)$?
$endgroup$
– LH8
Jan 22 at 20:02
$begingroup$
In general is true $f(U cap V) = f(U) cap f(V)$?
$endgroup$
– LH8
Jan 22 at 20:02
$begingroup$
If $f$ is not injective, $U cap V$ could be empty even if $f(U) cap f(V)$ is not.
$endgroup$
– Umberto P.
Jan 22 at 20:05
$begingroup$
If $f$ is not injective, $U cap V$ could be empty even if $f(U) cap f(V)$ is not.
$endgroup$
– Umberto P.
Jan 22 at 20:05
$begingroup$
And the converse? $f(U) cap f(V)$ not empty but $U cap V$ yes?
$endgroup$
– LH8
Jan 22 at 20:19
$begingroup$
And the converse? $f(U) cap f(V)$ not empty but $U cap V$ yes?
$endgroup$
– LH8
Jan 22 at 20:19
add a comment |
2 Answers
2
active
oldest
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$begingroup$
Let $f$ be continuous and suppose $y in f[overline{U}]$. So $y=f(x)$ with $x in overline{U}$. Now let $O$ be an open neighbourhood of $y$, then $f^{-1}[O]$ is an open neighbourhood of $x$ (here we use that $f$ is continuous), so $f^{-1}[O] cap U$ is non-empty (as $x in overline{U}$), say that $x' in f^{-1}[O] cap U$. But then $f(x') in f[U]$ and $f(x') in O$ so that $O$ intersects $f[U]$. As $O$ was an arbitrary open neighbourhood of $y$, $y in overline{f[U]}$ and the inclusion has been shown.
answered Jan 23 at 5:39
Henno BrandsmaHenno Brandsma
112k348120
$endgroup$
add a comment |
$begingroup$
You shouldn't have to use points at all. Since $overline{f(U)}$ is closed the continuity of $f$ implies $f^{-1}(overline{f(U)})$ is closed too. But $$U subset f^{-1}(f(U)) subset f^{-1}(overline{f(U)}) implies overline U subset f^{-1}(overline{f(U)})$$ because the latter set is closed. Thus
$$ f(overline U) subset f(f^{-1}(overline{f(U)})) = overline{f(U)}.$$
answered Jan 22 at 19:26
Umberto P.Umberto P.
39.8k13267
$endgroup$
$begingroup$
I know this proof, I´m interesting in the previous using elements and neighborhoods. Thanks.
$endgroup$
– LH8
Jan 22 at 19:48
$begingroup$
Please, answer my question using elements. I knew that proof.
$endgroup$
– LH8
Jan 22 at 20:47
add a comment |
Your Answer
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2 Answers
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2 Answers
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$begingroup$
Let $f$ be continuous and suppose $y in f[overline{U}]$. So $y=f(x)$ with $x in overline{U}$. Now let $O$ be an open neighbourhood of $y$, then $f^{-1}[O]$ is an open neighbourhood of $x$ (here we use that $f$ is continuous), so $f^{-1}[O] cap U$ is non-empty (as $x in overline{U}$), say that $x' in f^{-1}[O] cap U$. But then $f(x') in f[U]$ and $f(x') in O$ so that $O$ intersects $f[U]$. As $O$ was an arbitrary open neighbourhood of $y$, $y in overline{f[U]}$ and the inclusion has been shown.
answered Jan 23 at 5:39
Henno BrandsmaHenno Brandsma
112k348120
$endgroup$
add a comment |
$begingroup$
Let $f$ be continuous and suppose $y in f[overline{U}]$. So $y=f(x)$ with $x in overline{U}$. Now let $O$ be an open neighbourhood of $y$, then $f^{-1}[O]$ is an open neighbourhood of $x$ (here we use that $f$ is continuous), so $f^{-1}[O] cap U$ is non-empty (as $x in overline{U}$), say that $x' in f^{-1}[O] cap U$. But then $f(x') in f[U]$ and $f(x') in O$ so that $O$ intersects $f[U]$. As $O$ was an arbitrary open neighbourhood of $y$, $y in overline{f[U]}$ and the inclusion has been shown.
answered Jan 23 at 5:39
Henno BrandsmaHenno Brandsma
112k348120
$endgroup$
add a comment |
$begingroup$
Let $f$ be continuous and suppose $y in f[overline{U}]$. So $y=f(x)$ with $x in overline{U}$. Now let $O$ be an open neighbourhood of $y$, then $f^{-1}[O]$ is an open neighbourhood of $x$ (here we use that $f$ is continuous), so $f^{-1}[O] cap U$ is non-empty (as $x in overline{U}$), say that $x' in f^{-1}[O] cap U$. But then $f(x') in f[U]$ and $f(x') in O$ so that $O$ intersects $f[U]$. As $O$ was an arbitrary open neighbourhood of $y$, $y in overline{f[U]}$ and the inclusion has been shown.
answered Jan 23 at 5:39
Henno BrandsmaHenno Brandsma
112k348120
$endgroup$
Let $f$ be continuous and suppose $y in f[overline{U}]$. So $y=f(x)$ with $x in overline{U}$. Now let $O$ be an open neighbourhood of $y$, then $f^{-1}[O]$ is an open neighbourhood of $x$ (here we use that $f$ is continuous), so $f^{-1}[O] cap U$ is non-empty (as $x in overline{U}$), say that $x' in f^{-1}[O] cap U$. But then $f(x') in f[U]$ and $f(x') in O$ so that $O$ intersects $f[U]$. As $O$ was an arbitrary open neighbourhood of $y$, $y in overline{f[U]}$ and the inclusion has been shown.
answered Jan 23 at 5:39
Henno BrandsmaHenno Brandsma
112k348120
answered Jan 23 at 5:39
Henno BrandsmaHenno Brandsma
112k348120
answered Jan 23 at 5:39
Henno BrandsmaHenno Brandsma
112k348120
answered Jan 23 at 5:39
Henno BrandsmaHenno Brandsma
112k348120
112k348120
add a comment |
add a comment |
$begingroup$
You shouldn't have to use points at all. Since $overline{f(U)}$ is closed the continuity of $f$ implies $f^{-1}(overline{f(U)})$ is closed too. But $$U subset f^{-1}(f(U)) subset f^{-1}(overline{f(U)}) implies overline U subset f^{-1}(overline{f(U)})$$ because the latter set is closed. Thus
$$ f(overline U) subset f(f^{-1}(overline{f(U)})) = overline{f(U)}.$$
answered Jan 22 at 19:26
Umberto P.Umberto P.
39.8k13267
$endgroup$
$begingroup$
I know this proof, I´m interesting in the previous using elements and neighborhoods. Thanks.
$endgroup$
– LH8
Jan 22 at 19:48
$begingroup$
Please, answer my question using elements. I knew that proof.
$endgroup$
– LH8
Jan 22 at 20:47
add a comment |
$begingroup$
You shouldn't have to use points at all. Since $overline{f(U)}$ is closed the continuity of $f$ implies $f^{-1}(overline{f(U)})$ is closed too. But $$U subset f^{-1}(f(U)) subset f^{-1}(overline{f(U)}) implies overline U subset f^{-1}(overline{f(U)})$$ because the latter set is closed. Thus
$$ f(overline U) subset f(f^{-1}(overline{f(U)})) = overline{f(U)}.$$
answered Jan 22 at 19:26
Umberto P.Umberto P.
39.8k13267
$endgroup$
$begingroup$
I know this proof, I´m interesting in the previous using elements and neighborhoods. Thanks.
$endgroup$
– LH8
Jan 22 at 19:48
$begingroup$
Please, answer my question using elements. I knew that proof.
$endgroup$
– LH8
Jan 22 at 20:47
add a comment |
$begingroup$
You shouldn't have to use points at all. Since $overline{f(U)}$ is closed the continuity of $f$ implies $f^{-1}(overline{f(U)})$ is closed too. But $$U subset f^{-1}(f(U)) subset f^{-1}(overline{f(U)}) implies overline U subset f^{-1}(overline{f(U)})$$ because the latter set is closed. Thus
$$ f(overline U) subset f(f^{-1}(overline{f(U)})) = overline{f(U)}.$$
answered Jan 22 at 19:26
Umberto P.Umberto P.
39.8k13267
$endgroup$
You shouldn't have to use points at all. Since $overline{f(U)}$ is closed the continuity of $f$ implies $f^{-1}(overline{f(U)})$ is closed too. But $$U subset f^{-1}(f(U)) subset f^{-1}(overline{f(U)}) implies overline U subset f^{-1}(overline{f(U)})$$ because the latter set is closed. Thus
$$ f(overline U) subset f(f^{-1}(overline{f(U)})) = overline{f(U)}.$$
answered Jan 22 at 19:26
Umberto P.Umberto P.
39.8k13267
answered Jan 22 at 19:26
Umberto P.Umberto P.
39.8k13267
answered Jan 22 at 19:26
Umberto P.Umberto P.
39.8k13267
answered Jan 22 at 19:26
Umberto P.Umberto P.
39.8k13267
39.8k13267
$begingroup$
I know this proof, I´m interesting in the previous using elements and neighborhoods. Thanks.
$endgroup$
– LH8
Jan 22 at 19:48
$begingroup$
Please, answer my question using elements. I knew that proof.
$endgroup$
– LH8
Jan 22 at 20:47
add a comment |
$begingroup$
I know this proof, I´m interesting in the previous using elements and neighborhoods. Thanks.
$endgroup$
– LH8
Jan 22 at 19:48
$begingroup$
Please, answer my question using elements. I knew that proof.
$endgroup$
– LH8
Jan 22 at 20:47
$begingroup$
I know this proof, I´m interesting in the previous using elements and neighborhoods. Thanks.
$endgroup$
– LH8
Jan 22 at 19:48
$begingroup$
I know this proof, I´m interesting in the previous using elements and neighborhoods. Thanks.
$endgroup$
– LH8
Jan 22 at 19:48
$begingroup$
Please, answer my question using elements. I knew that proof.
$endgroup$
– LH8
Jan 22 at 20:47
$begingroup$
Please, answer my question using elements. I knew that proof.
$endgroup$
– LH8
Jan 22 at 20:47
add a comment |
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$begingroup$
Be careful about which space your neighborhoods are in. You write "let $xin f(overline U)$. So for any open neigborhood $V$ of $x$ is $Vcap f(U)neq varnothing$". This is not right. What you really have is that $x = f(x')$ for some $xin X$ such that for every open neigborhood $Vni x$, $Vcap Uneqvarnothing$. EDIT: Sorry, when you said "so [etc]", I thought you claimed that [etc] was a consequence.
$endgroup$
– o.h.
Jan 22 at 19:25
$begingroup$
In general is true $f(U cap V) = f(U) cap f(V)$?
$endgroup$
– LH8
Jan 22 at 20:02
$begingroup$
If $f$ is not injective, $U cap V$ could be empty even if $f(U) cap f(V)$ is not.
$endgroup$
– Umberto P.
Jan 22 at 20:05
$begingroup$
And the converse? $f(U) cap f(V)$ not empty but $U cap V$ yes?
$endgroup$
– LH8
Jan 22 at 20:19