Mixing
Why, in topology, is continuity defined with inverse images? [duplicate]
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This question already has an answer here:
Why aren’t continuous functions defined the other way around?
6 answers
In topology we defined a continuous map to be a map between 2 spaces $f: X mapsto Y$ such that if $Usubset Y$ is open then $f^{-1}(U)$ is open.
Why did we use the preimage; why not say 'a continuous map always maps open sets to open sets'?
general-topology
edited Feb 3 at 2:02
J. W. Tanner
4,7871420
asked Feb 2 at 19:54
Toby PeterkenToby Peterken
1496
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marked as duplicate by Henno Brandsma general-topology
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Feb 2 at 20:25
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Why aren’t continuous functions defined the other way around?
6 answers
In topology we defined a continuous map to be a map between 2 spaces $f: X mapsto Y$ such that if $Usubset Y$ is open then $f^{-1}(U)$ is open.
Why did we use the preimage; why not say 'a continuous map always maps open sets to open sets'?
general-topology
edited Feb 3 at 2:02
J. W. Tanner
4,7871420
asked Feb 2 at 19:54
Toby PeterkenToby Peterken
1496
$endgroup$
marked as duplicate by Henno Brandsma general-topology
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Feb 2 at 20:25
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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Because then the definition would disagree with the one we use in real analysis; not all open maps are continuous there.
$endgroup$
– dbx
Feb 2 at 19:56
1
$begingroup$
Do you want $f: mathbb{R} to Bbb R : f(x)=x^2$ to be continuous? It sends the open set $(-1,1)$ to the non-open $[0,1)$ but it does obey the inverse image definition.
$endgroup$
– Henno Brandsma
Feb 2 at 20:02
$begingroup$
Constant functions will also usually be continuous but not open.
$endgroup$
– Jon Hillery
Feb 2 at 20:04
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I think the answer to this question is relevant here.
$endgroup$
– kimchi lover
Feb 2 at 20:05
$begingroup$
"Why did we use the preimage, why not say ' a coninuous map always maps open sets to open sets'?" Because they don't. $f(x) = 2$ maps $(0,10)$ to ${2}$ and $f(x) = sin x$ maps $(0, 10)$ to $[-1,1]$ neither of which are open. More to the point an open set can only have open sets mapped to it.
$endgroup$
– fleablood
Feb 2 at 20:23
add a comment |
$begingroup$
This question already has an answer here:
Why aren’t continuous functions defined the other way around?
6 answers
In topology we defined a continuous map to be a map between 2 spaces $f: X mapsto Y$ such that if $Usubset Y$ is open then $f^{-1}(U)$ is open.
Why did we use the preimage; why not say 'a continuous map always maps open sets to open sets'?
general-topology
edited Feb 3 at 2:02
J. W. Tanner
4,7871420
asked Feb 2 at 19:54
Toby PeterkenToby Peterken
1496
$endgroup$
This question already has an answer here:
Why aren’t continuous functions defined the other way around?
6 answers
In topology we defined a continuous map to be a map between 2 spaces $f: X mapsto Y$ such that if $Usubset Y$ is open then $f^{-1}(U)$ is open.
Why did we use the preimage; why not say 'a continuous map always maps open sets to open sets'?
This question already has an answer here:
Why aren’t continuous functions defined the other way around?
6 answers
general-topology
general-topology
edited Feb 3 at 2:02
J. W. Tanner
4,7871420
asked Feb 2 at 19:54
Toby PeterkenToby Peterken
1496
edited Feb 3 at 2:02
J. W. Tanner
4,7871420
asked Feb 2 at 19:54
Toby PeterkenToby Peterken
1496
edited Feb 3 at 2:02
J. W. Tanner
4,7871420
edited Feb 3 at 2:02
J. W. Tanner
4,7871420
edited Feb 3 at 2:02
J. W. Tanner
4,7871420
4,7871420
asked Feb 2 at 19:54
Toby PeterkenToby Peterken
1496
asked Feb 2 at 19:54
Toby PeterkenToby Peterken
1496
asked Feb 2 at 19:54
Toby PeterkenToby Peterken
1496
1496
marked as duplicate by Henno Brandsma general-topology
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Feb 2 at 20:25
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Henno Brandsma general-topology
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Feb 2 at 20:25
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
Because then the definition would disagree with the one we use in real analysis; not all open maps are continuous there.
$endgroup$
– dbx
Feb 2 at 19:56
1
$begingroup$
Do you want $f: mathbb{R} to Bbb R : f(x)=x^2$ to be continuous? It sends the open set $(-1,1)$ to the non-open $[0,1)$ but it does obey the inverse image definition.
$endgroup$
– Henno Brandsma
Feb 2 at 20:02
$begingroup$
Constant functions will also usually be continuous but not open.
$endgroup$
– Jon Hillery
Feb 2 at 20:04
$begingroup$
I think the answer to this question is relevant here.
$endgroup$
– kimchi lover
Feb 2 at 20:05
$begingroup$
"Why did we use the preimage, why not say ' a coninuous map always maps open sets to open sets'?" Because they don't. $f(x) = 2$ maps $(0,10)$ to ${2}$ and $f(x) = sin x$ maps $(0, 10)$ to $[-1,1]$ neither of which are open. More to the point an open set can only have open sets mapped to it.
$endgroup$
– fleablood
Feb 2 at 20:23
add a comment |
$begingroup$
Because then the definition would disagree with the one we use in real analysis; not all open maps are continuous there.
$endgroup$
– dbx
Feb 2 at 19:56
1
$begingroup$
Do you want $f: mathbb{R} to Bbb R : f(x)=x^2$ to be continuous? It sends the open set $(-1,1)$ to the non-open $[0,1)$ but it does obey the inverse image definition.
$endgroup$
– Henno Brandsma
Feb 2 at 20:02
$begingroup$
Constant functions will also usually be continuous but not open.
$endgroup$
– Jon Hillery
Feb 2 at 20:04
$begingroup$
I think the answer to this question is relevant here.
$endgroup$
– kimchi lover
Feb 2 at 20:05
$begingroup$
"Why did we use the preimage, why not say ' a coninuous map always maps open sets to open sets'?" Because they don't. $f(x) = 2$ maps $(0,10)$ to ${2}$ and $f(x) = sin x$ maps $(0, 10)$ to $[-1,1]$ neither of which are open. More to the point an open set can only have open sets mapped to it.
$endgroup$
– fleablood
Feb 2 at 20:23
$begingroup$
Because then the definition would disagree with the one we use in real analysis; not all open maps are continuous there.
$endgroup$
– dbx
Feb 2 at 19:56
$begingroup$
Because then the definition would disagree with the one we use in real analysis; not all open maps are continuous there.
$endgroup$
– dbx
Feb 2 at 19:56
1
1
$begingroup$
Do you want $f: mathbb{R} to Bbb R : f(x)=x^2$ to be continuous? It sends the open set $(-1,1)$ to the non-open $[0,1)$ but it does obey the inverse image definition.
$endgroup$
– Henno Brandsma
Feb 2 at 20:02
$begingroup$
Do you want $f: mathbb{R} to Bbb R : f(x)=x^2$ to be continuous? It sends the open set $(-1,1)$ to the non-open $[0,1)$ but it does obey the inverse image definition.
$endgroup$
– Henno Brandsma
Feb 2 at 20:02
$begingroup$
Constant functions will also usually be continuous but not open.
$endgroup$
– Jon Hillery
Feb 2 at 20:04
$begingroup$
Constant functions will also usually be continuous but not open.
$endgroup$
– Jon Hillery
Feb 2 at 20:04
$begingroup$
I think the answer to this question is relevant here.
$endgroup$
– kimchi lover
Feb 2 at 20:05
$begingroup$
I think the answer to this question is relevant here.
$endgroup$
– kimchi lover
Feb 2 at 20:05
$begingroup$
"Why did we use the preimage, why not say ' a coninuous map always maps open sets to open sets'?" Because they don't. $f(x) = 2$ maps $(0,10)$ to ${2}$ and $f(x) = sin x$ maps $(0, 10)$ to $[-1,1]$ neither of which are open. More to the point an open set can only have open sets mapped to it.
$endgroup$
– fleablood
Feb 2 at 20:23
$begingroup$
"Why did we use the preimage, why not say ' a coninuous map always maps open sets to open sets'?" Because they don't. $f(x) = 2$ maps $(0,10)$ to ${2}$ and $f(x) = sin x$ maps $(0, 10)$ to $[-1,1]$ neither of which are open. More to the point an open set can only have open sets mapped to it.
$endgroup$
– fleablood
Feb 2 at 20:23
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Because with your definition we would have to agree that the function $f:mathbb{R}tomathbb{R}$ defined by $f(x)=0$ (constant function) is not continuous in the standard topology on $mathbb{R}$. We don't want that to happen.
Now, in metric spaces we can define continuity with $epsilon-delta$ like it is being done in analysis. But then there is a theorem that a function $f:(X,d_1)to (Y,d_2)$ between metric spaces is continuous if and only if for each open set $Usubset Y$ the set $f^{-1}(U)subset X$ is open. You can take it as a good exercise to prove this theorem. And this is the motivation for the definition of continuity in topological spaces which are not necessary metric spaces. With this definition we get a generalization.
answered Feb 2 at 20:03
MarkMark
10.5k1622
$endgroup$
add a comment |
$begingroup$
The statement "a coninuous map always maps open sets to open sets" is false. A counter example is $f(x)=x^2$ on $mathbb R rightarrow mathbb R$ . Then the open set $(-1,1)$ is sent to the non-open set $[0,1)$.
answered Feb 2 at 20:04
Picaud VincentPicaud Vincent
1,434310
$endgroup$
$begingroup$
For additional fun, the same map, but viewed as $Bbb Rto[0,infty)$ would be continuous in the proposed sense. (But that trick does not work for $xmapsto x^3-x$.)
$endgroup$
– Hagen von Eitzen
Feb 2 at 20:05
$begingroup$
@HagenvonEitzen So let's take $sin(x)$ that maps $(-10,10)$ to $[-1,1]$ :-)
$endgroup$
– Picaud Vincent
Feb 2 at 20:07
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Because with your definition we would have to agree that the function $f:mathbb{R}tomathbb{R}$ defined by $f(x)=0$ (constant function) is not continuous in the standard topology on $mathbb{R}$. We don't want that to happen.
Now, in metric spaces we can define continuity with $epsilon-delta$ like it is being done in analysis. But then there is a theorem that a function $f:(X,d_1)to (Y,d_2)$ between metric spaces is continuous if and only if for each open set $Usubset Y$ the set $f^{-1}(U)subset X$ is open. You can take it as a good exercise to prove this theorem. And this is the motivation for the definition of continuity in topological spaces which are not necessary metric spaces. With this definition we get a generalization.
answered Feb 2 at 20:03
MarkMark
10.5k1622
$endgroup$
add a comment |
$begingroup$
Because with your definition we would have to agree that the function $f:mathbb{R}tomathbb{R}$ defined by $f(x)=0$ (constant function) is not continuous in the standard topology on $mathbb{R}$. We don't want that to happen.
Now, in metric spaces we can define continuity with $epsilon-delta$ like it is being done in analysis. But then there is a theorem that a function $f:(X,d_1)to (Y,d_2)$ between metric spaces is continuous if and only if for each open set $Usubset Y$ the set $f^{-1}(U)subset X$ is open. You can take it as a good exercise to prove this theorem. And this is the motivation for the definition of continuity in topological spaces which are not necessary metric spaces. With this definition we get a generalization.
answered Feb 2 at 20:03
MarkMark
10.5k1622
$endgroup$
add a comment |
$begingroup$
Because with your definition we would have to agree that the function $f:mathbb{R}tomathbb{R}$ defined by $f(x)=0$ (constant function) is not continuous in the standard topology on $mathbb{R}$. We don't want that to happen.
Now, in metric spaces we can define continuity with $epsilon-delta$ like it is being done in analysis. But then there is a theorem that a function $f:(X,d_1)to (Y,d_2)$ between metric spaces is continuous if and only if for each open set $Usubset Y$ the set $f^{-1}(U)subset X$ is open. You can take it as a good exercise to prove this theorem. And this is the motivation for the definition of continuity in topological spaces which are not necessary metric spaces. With this definition we get a generalization.
answered Feb 2 at 20:03
MarkMark
10.5k1622
$endgroup$
Because with your definition we would have to agree that the function $f:mathbb{R}tomathbb{R}$ defined by $f(x)=0$ (constant function) is not continuous in the standard topology on $mathbb{R}$. We don't want that to happen.
Now, in metric spaces we can define continuity with $epsilon-delta$ like it is being done in analysis. But then there is a theorem that a function $f:(X,d_1)to (Y,d_2)$ between metric spaces is continuous if and only if for each open set $Usubset Y$ the set $f^{-1}(U)subset X$ is open. You can take it as a good exercise to prove this theorem. And this is the motivation for the definition of continuity in topological spaces which are not necessary metric spaces. With this definition we get a generalization.
answered Feb 2 at 20:03
MarkMark
10.5k1622
answered Feb 2 at 20:03
MarkMark
10.5k1622
answered Feb 2 at 20:03
MarkMark
10.5k1622
answered Feb 2 at 20:03
MarkMark
10.5k1622
10.5k1622
add a comment |
add a comment |
$begingroup$
The statement "a coninuous map always maps open sets to open sets" is false. A counter example is $f(x)=x^2$ on $mathbb R rightarrow mathbb R$ . Then the open set $(-1,1)$ is sent to the non-open set $[0,1)$.
answered Feb 2 at 20:04
Picaud VincentPicaud Vincent
1,434310
$endgroup$
$begingroup$
For additional fun, the same map, but viewed as $Bbb Rto[0,infty)$ would be continuous in the proposed sense. (But that trick does not work for $xmapsto x^3-x$.)
$endgroup$
– Hagen von Eitzen
Feb 2 at 20:05
$begingroup$
@HagenvonEitzen So let's take $sin(x)$ that maps $(-10,10)$ to $[-1,1]$ :-)
$endgroup$
– Picaud Vincent
Feb 2 at 20:07
add a comment |
$begingroup$
The statement "a coninuous map always maps open sets to open sets" is false. A counter example is $f(x)=x^2$ on $mathbb R rightarrow mathbb R$ . Then the open set $(-1,1)$ is sent to the non-open set $[0,1)$.
answered Feb 2 at 20:04
Picaud VincentPicaud Vincent
1,434310
$endgroup$
$begingroup$
For additional fun, the same map, but viewed as $Bbb Rto[0,infty)$ would be continuous in the proposed sense. (But that trick does not work for $xmapsto x^3-x$.)
$endgroup$
– Hagen von Eitzen
Feb 2 at 20:05
$begingroup$
@HagenvonEitzen So let's take $sin(x)$ that maps $(-10,10)$ to $[-1,1]$ :-)
$endgroup$
– Picaud Vincent
Feb 2 at 20:07
add a comment |
$begingroup$
The statement "a coninuous map always maps open sets to open sets" is false. A counter example is $f(x)=x^2$ on $mathbb R rightarrow mathbb R$ . Then the open set $(-1,1)$ is sent to the non-open set $[0,1)$.
answered Feb 2 at 20:04
Picaud VincentPicaud Vincent
1,434310
$endgroup$
The statement "a coninuous map always maps open sets to open sets" is false. A counter example is $f(x)=x^2$ on $mathbb R rightarrow mathbb R$ . Then the open set $(-1,1)$ is sent to the non-open set $[0,1)$.
answered Feb 2 at 20:04
Picaud VincentPicaud Vincent
1,434310
answered Feb 2 at 20:04
Picaud VincentPicaud Vincent
1,434310
answered Feb 2 at 20:04
Picaud VincentPicaud Vincent
1,434310
answered Feb 2 at 20:04
Picaud VincentPicaud Vincent
1,434310
1,434310
$begingroup$
For additional fun, the same map, but viewed as $Bbb Rto[0,infty)$ would be continuous in the proposed sense. (But that trick does not work for $xmapsto x^3-x$.)
$endgroup$
– Hagen von Eitzen
Feb 2 at 20:05
$begingroup$
@HagenvonEitzen So let's take $sin(x)$ that maps $(-10,10)$ to $[-1,1]$ :-)
$endgroup$
– Picaud Vincent
Feb 2 at 20:07
add a comment |
$begingroup$
For additional fun, the same map, but viewed as $Bbb Rto[0,infty)$ would be continuous in the proposed sense. (But that trick does not work for $xmapsto x^3-x$.)
$endgroup$
– Hagen von Eitzen
Feb 2 at 20:05
$begingroup$
@HagenvonEitzen So let's take $sin(x)$ that maps $(-10,10)$ to $[-1,1]$ :-)
$endgroup$
– Picaud Vincent
Feb 2 at 20:07
$begingroup$
For additional fun, the same map, but viewed as $Bbb Rto[0,infty)$ would be continuous in the proposed sense. (But that trick does not work for $xmapsto x^3-x$.)
$endgroup$
– Hagen von Eitzen
Feb 2 at 20:05
$begingroup$
For additional fun, the same map, but viewed as $Bbb Rto[0,infty)$ would be continuous in the proposed sense. (But that trick does not work for $xmapsto x^3-x$.)
$endgroup$
– Hagen von Eitzen
Feb 2 at 20:05
$begingroup$
@HagenvonEitzen So let's take $sin(x)$ that maps $(-10,10)$ to $[-1,1]$ :-)
$endgroup$
– Picaud Vincent
Feb 2 at 20:07
$begingroup$
@HagenvonEitzen So let's take $sin(x)$ that maps $(-10,10)$ to $[-1,1]$ :-)
$endgroup$
– Picaud Vincent
Feb 2 at 20:07
add a comment |
$begingroup$
Because then the definition would disagree with the one we use in real analysis; not all open maps are continuous there.
$endgroup$
– dbx
Feb 2 at 19:56
1
$begingroup$
Do you want $f: mathbb{R} to Bbb R : f(x)=x^2$ to be continuous? It sends the open set $(-1,1)$ to the non-open $[0,1)$ but it does obey the inverse image definition.
$endgroup$
– Henno Brandsma
Feb 2 at 20:02
$begingroup$
Constant functions will also usually be continuous but not open.
$endgroup$
– Jon Hillery
Feb 2 at 20:04
$begingroup$
I think the answer to this question is relevant here.
$endgroup$
– kimchi lover
Feb 2 at 20:05
$begingroup$
"Why did we use the preimage, why not say ' a coninuous map always maps open sets to open sets'?" Because they don't. $f(x) = 2$ maps $(0,10)$ to ${2}$ and $f(x) = sin x$ maps $(0, 10)$ to $[-1,1]$ neither of which are open. More to the point an open set can only have open sets mapped to it.
$endgroup$
– fleablood
Feb 2 at 20:23