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How to prove or disprove the following statement on Hausdorff topology

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0

$begingroup$

Let $X$ be a set and let $Y$ be a Hausdorff space. Let $fcolon X to Y$ be a given mapping. Define $U subset X $ to be open in $X$ if, and only if $U = f^{-1}(V)$ for some set $V$ open in $Y$. This defines a Hausdorff topology on $X$.

I have no idea how to start with.

general-topology weak-topology

share|cite|improve this question

asked Mar 19 at 16:02

MöbiusMöbius

367

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I presume you know that a space is Hausdorff if and only if, given two points, p and q, there exist disjoint sets U and V such that p is in U and q is in V. Here we are given a Hausdorff space y, and a function, f, from X to Y.
  $endgroup$
  – user247327
  Mar 19 at 16:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You need $f$ to be injective for the topology to be Hausdorff.
  $endgroup$
  – Robert Thingum
  Mar 19 at 16:10
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $X$ be a set and let $Y$ be a Hausdorff space. Let $fcolon X to Y$ be a given mapping. Define $U subset X $ to be open in $X$ if, and only if $U = f^{-1}(V)$ for some set $V$ open in $Y$. This defines a Hausdorff topology on $X$.

I have no idea how to start with.

general-topology weak-topology

share|cite|improve this question

asked Mar 19 at 16:02

MöbiusMöbius

367

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I presume you know that a space is Hausdorff if and only if, given two points, p and q, there exist disjoint sets U and V such that p is in U and q is in V. Here we are given a Hausdorff space y, and a function, f, from X to Y.
  $endgroup$
  – user247327
  Mar 19 at 16:07
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You need $f$ to be injective for the topology to be Hausdorff.
  $endgroup$
  – Robert Thingum
  Mar 19 at 16:10
  

  
  
  
  

add a comment | 

$begingroup$

Let $X$ be a set and let $Y$ be a Hausdorff space. Let $fcolon X to Y$ be a given mapping. Define $U subset X $ to be open in $X$ if, and only if $U = f^{-1}(V)$ for some set $V$ open in $Y$. This defines a Hausdorff topology on $X$.

I have no idea how to start with.

general-topology weak-topology

share|cite|improve this question

asked Mar 19 at 16:02

MöbiusMöbius

367

$endgroup$

share|cite|improve this question

asked Mar 19 at 16:02

MöbiusMöbius

367

share|cite|improve this question

asked Mar 19 at 16:02

MöbiusMöbius

367

asked Mar 19 at 16:02

MöbiusMöbius

367

asked Mar 19 at 16:02

MöbiusMöbius

367

3 Answers
3

active

oldest

votes

1

$begingroup$

There's nowhere else to start, but with definitions! That having been said, I'm assuming you know the definition of a topology.

Let $tau$ be the collection of sets described in your question. That is, $tau$ is the set of all $Usubseteq X$ such that $U=f^{-1}(V)$ for some open $Vsubseteq Y$. Then,

$$X=f^{-1}(Y)$$

so $Xintau$.

$$emptyset=f^{-1}(emptyset)$$

so $emptysetintau$.

Let $U_{1},ldots,U_{n}intau$, say $U_{i}=f^{-1}(V_{i})$ where $V_{i}subseteq Y$ is open. Then,

$$bigcap_{1}^{n}U_{i}=bigcap_{1}^{n}f^{-1}(V_{i})=f^{-1}left(bigcap_{1}^{n}V_{i}right)$$

Certainly $bigcap_{1}^{n}V_{i}$ is an open set of $Y$, so $bigcap_{1}^{n}U_{i}$ is an element of $tau$. Next we let ${U_{alpha}}$ be a collection of elements of $tau$, say $U_{alpha}=f^{-1}(V_{alpha})$ where $V_{alpha}$ is open in $Y$. Then

$$bigcup_{alpha}U_{alpha}=bigcup_{alpha}f^{-1}(V_{alpha})=f^{-1}left(bigcup_{alpha}V_{alpha}right)$$

Certainly $bigcup_{alpha}V_{alpha}$ is open in $Y$, so $bigcup_{alpha}U_{alpha}$ is an element of $tau$. We then have that $tau$ is a topology on $X$.

This topology is not necessarily Hausdorff

Let $X$ be the underlying set of $mathbb{R}$ and let $Y={0}$ with the indiscrete topology. Then $Y$ is Hausdorff, and the constant function $f:Xrightarrow Y$ can be our given map. The topology described above is then the indiscrete topology on $X$, which is not Hausdorff.

If the function is injective, then the topology is Hausdorff.

If $f$ is injective, then given distinct $x,yin X$ we have that $f(x)$ and $f(y)$ are distinct in $Y$. Then, because $Y$ is Hausdorff there would be disjoint open sets $V_{1},V_{2}subseteq Y$ containing $f(x)$ and $f(y),$ respectively. Then $f^{-1}(V_{1})$ and $f^{-1}(V_{2})$ are disjoint open sets of $X$ that contain $x$ and $y$, respectively. Thus $X$ equipped with $tau$ is Hausdorff.

share|cite|improve this answer

edited Mar 19 at 16:27

answered Mar 19 at 16:10

Robert ThingumRobert Thingum

9711317

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• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much. I am new in Topology and try to understand all it's definitions properly.
  $endgroup$
  – Möbius
  Mar 19 at 16:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Robert.Very nice.
  $endgroup$
  – Peter Szilas
  Mar 19 at 17:50
  

  
  
  
  

add a comment | 

1

$begingroup$

Let's suppose that $f$ is injective.

Let $x_1 neq x_2 in X$. Consider their images $f(x_1)$ and $f(x_2)$. They are different because $f$ is injective. Because $Y$ is Hausdorff, there exists two open subsets $U_1$ and $U_2$ of $Y$, with $f(x_1) in U_1$ and $f(x_2) in U_2$, and $U_1 cap U_2 = emptyset$.

Now $f^{-1}(U_1)$ and $f^{-1}(U_2)$ are open in $X$ by definition of open sets on $X$. They contain respectively $x_1$ and $x_2$. And they don't intersect, otherwise $U_1$ et $U_2$ would intersect.

This shows that $X$ is Hausdorff.

share|cite|improve this answer

edited Mar 19 at 16:10

answered Mar 19 at 16:07

TheSilverDoeTheSilverDoe

5,433216

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What if $f(x_{1})=f(x_{2})$?
  $endgroup$
  – Robert Thingum
  Mar 19 at 16:08
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for the precision. Indeed the property does not hold in general. I add that I suppose that $f$ is injective.
  $endgroup$
  – TheSilverDoe
  Mar 19 at 16:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  f:X-> Y is a "given mapping". What is the definition of "mapping" in topology?
  $endgroup$
  – user247327
  Mar 19 at 16:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user247327 What do you mean ? The property the OP wants to show is not true in general. I just give a proof when you add the asumption that $f$ is injective, in which case the property holds.
  $endgroup$
  – TheSilverDoe
  Mar 19 at 16:18
  

  
  
  
  

add a comment | 

0

$begingroup$

This is not true in general. Suppose $f(x)=f(y)$, then for all $Usubset X$ open with $xin U$ there exists a $Vsubset Y$ open such that $U=f^{-1}(V)$. Note that $xin U$ if and only if $f(x)=f(y)in V$ which implies that $yin U$. So you need that $f$ is injective. If that holds you can follow the answer of TheSilverDoe.

share|cite|improve this answer

answered Mar 19 at 16:09

Floris ClaassensFloris Claassens

1,33229

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add a comment | 

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1

$begingroup$

There's nowhere else to start, but with definitions! That having been said, I'm assuming you know the definition of a topology.

Let $tau$ be the collection of sets described in your question. That is, $tau$ is the set of all $Usubseteq X$ such that $U=f^{-1}(V)$ for some open $Vsubseteq Y$. Then,

$$X=f^{-1}(Y)$$

so $Xintau$.

$$emptyset=f^{-1}(emptyset)$$

so $emptysetintau$.

Let $U_{1},ldots,U_{n}intau$, say $U_{i}=f^{-1}(V_{i})$ where $V_{i}subseteq Y$ is open. Then,

$$bigcap_{1}^{n}U_{i}=bigcap_{1}^{n}f^{-1}(V_{i})=f^{-1}left(bigcap_{1}^{n}V_{i}right)$$

Certainly $bigcap_{1}^{n}V_{i}$ is an open set of $Y$, so $bigcap_{1}^{n}U_{i}$ is an element of $tau$. Next we let ${U_{alpha}}$ be a collection of elements of $tau$, say $U_{alpha}=f^{-1}(V_{alpha})$ where $V_{alpha}$ is open in $Y$. Then

$$bigcup_{alpha}U_{alpha}=bigcup_{alpha}f^{-1}(V_{alpha})=f^{-1}left(bigcup_{alpha}V_{alpha}right)$$

Certainly $bigcup_{alpha}V_{alpha}$ is open in $Y$, so $bigcup_{alpha}U_{alpha}$ is an element of $tau$. We then have that $tau$ is a topology on $X$.

This topology is not necessarily Hausdorff

Let $X$ be the underlying set of $mathbb{R}$ and let $Y={0}$ with the indiscrete topology. Then $Y$ is Hausdorff, and the constant function $f:Xrightarrow Y$ can be our given map. The topology described above is then the indiscrete topology on $X$, which is not Hausdorff.

If the function is injective, then the topology is Hausdorff.

If $f$ is injective, then given distinct $x,yin X$ we have that $f(x)$ and $f(y)$ are distinct in $Y$. Then, because $Y$ is Hausdorff there would be disjoint open sets $V_{1},V_{2}subseteq Y$ containing $f(x)$ and $f(y),$ respectively. Then $f^{-1}(V_{1})$ and $f^{-1}(V_{2})$ are disjoint open sets of $X$ that contain $x$ and $y$, respectively. Thus $X$ equipped with $tau$ is Hausdorff.

share|cite|improve this answer

edited Mar 19 at 16:27

answered Mar 19 at 16:10

Robert ThingumRobert Thingum

9711317

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much. I am new in Topology and try to understand all it's definitions properly.
  $endgroup$
  – Möbius
  Mar 19 at 16:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Robert.Very nice.
  $endgroup$
  – Peter Szilas
  Mar 19 at 17:50
  

  
  
  
  

add a comment | 

1

$begingroup$

There's nowhere else to start, but with definitions! That having been said, I'm assuming you know the definition of a topology.

Let $tau$ be the collection of sets described in your question. That is, $tau$ is the set of all $Usubseteq X$ such that $U=f^{-1}(V)$ for some open $Vsubseteq Y$. Then,

$$X=f^{-1}(Y)$$

so $Xintau$.

$$emptyset=f^{-1}(emptyset)$$

so $emptysetintau$.

Let $U_{1},ldots,U_{n}intau$, say $U_{i}=f^{-1}(V_{i})$ where $V_{i}subseteq Y$ is open. Then,

$$bigcap_{1}^{n}U_{i}=bigcap_{1}^{n}f^{-1}(V_{i})=f^{-1}left(bigcap_{1}^{n}V_{i}right)$$

Certainly $bigcap_{1}^{n}V_{i}$ is an open set of $Y$, so $bigcap_{1}^{n}U_{i}$ is an element of $tau$. Next we let ${U_{alpha}}$ be a collection of elements of $tau$, say $U_{alpha}=f^{-1}(V_{alpha})$ where $V_{alpha}$ is open in $Y$. Then

$$bigcup_{alpha}U_{alpha}=bigcup_{alpha}f^{-1}(V_{alpha})=f^{-1}left(bigcup_{alpha}V_{alpha}right)$$

Certainly $bigcup_{alpha}V_{alpha}$ is open in $Y$, so $bigcup_{alpha}U_{alpha}$ is an element of $tau$. We then have that $tau$ is a topology on $X$.

This topology is not necessarily Hausdorff

Let $X$ be the underlying set of $mathbb{R}$ and let $Y={0}$ with the indiscrete topology. Then $Y$ is Hausdorff, and the constant function $f:Xrightarrow Y$ can be our given map. The topology described above is then the indiscrete topology on $X$, which is not Hausdorff.

If the function is injective, then the topology is Hausdorff.

If $f$ is injective, then given distinct $x,yin X$ we have that $f(x)$ and $f(y)$ are distinct in $Y$. Then, because $Y$ is Hausdorff there would be disjoint open sets $V_{1},V_{2}subseteq Y$ containing $f(x)$ and $f(y),$ respectively. Then $f^{-1}(V_{1})$ and $f^{-1}(V_{2})$ are disjoint open sets of $X$ that contain $x$ and $y$, respectively. Thus $X$ equipped with $tau$ is Hausdorff.

share|cite|improve this answer

edited Mar 19 at 16:27

answered Mar 19 at 16:10

Robert ThingumRobert Thingum

9711317

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much. I am new in Topology and try to understand all it's definitions properly.
  $endgroup$
  – Möbius
  Mar 19 at 16:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Robert.Very nice.
  $endgroup$
  – Peter Szilas
  Mar 19 at 17:50
  

  
  
  
  

add a comment | 

$begingroup$

There's nowhere else to start, but with definitions! That having been said, I'm assuming you know the definition of a topology.

Let $tau$ be the collection of sets described in your question. That is, $tau$ is the set of all $Usubseteq X$ such that $U=f^{-1}(V)$ for some open $Vsubseteq Y$. Then,

$$X=f^{-1}(Y)$$

so $Xintau$.

$$emptyset=f^{-1}(emptyset)$$

so $emptysetintau$.

Let $U_{1},ldots,U_{n}intau$, say $U_{i}=f^{-1}(V_{i})$ where $V_{i}subseteq Y$ is open. Then,

$$bigcap_{1}^{n}U_{i}=bigcap_{1}^{n}f^{-1}(V_{i})=f^{-1}left(bigcap_{1}^{n}V_{i}right)$$

Certainly $bigcap_{1}^{n}V_{i}$ is an open set of $Y$, so $bigcap_{1}^{n}U_{i}$ is an element of $tau$. Next we let ${U_{alpha}}$ be a collection of elements of $tau$, say $U_{alpha}=f^{-1}(V_{alpha})$ where $V_{alpha}$ is open in $Y$. Then

$$bigcup_{alpha}U_{alpha}=bigcup_{alpha}f^{-1}(V_{alpha})=f^{-1}left(bigcup_{alpha}V_{alpha}right)$$

Certainly $bigcup_{alpha}V_{alpha}$ is open in $Y$, so $bigcup_{alpha}U_{alpha}$ is an element of $tau$. We then have that $tau$ is a topology on $X$.

This topology is not necessarily Hausdorff

Let $X$ be the underlying set of $mathbb{R}$ and let $Y={0}$ with the indiscrete topology. Then $Y$ is Hausdorff, and the constant function $f:Xrightarrow Y$ can be our given map. The topology described above is then the indiscrete topology on $X$, which is not Hausdorff.

If the function is injective, then the topology is Hausdorff.

If $f$ is injective, then given distinct $x,yin X$ we have that $f(x)$ and $f(y)$ are distinct in $Y$. Then, because $Y$ is Hausdorff there would be disjoint open sets $V_{1},V_{2}subseteq Y$ containing $f(x)$ and $f(y),$ respectively. Then $f^{-1}(V_{1})$ and $f^{-1}(V_{2})$ are disjoint open sets of $X$ that contain $x$ and $y$, respectively. Thus $X$ equipped with $tau$ is Hausdorff.

share|cite|improve this answer

edited Mar 19 at 16:27

answered Mar 19 at 16:10

Robert ThingumRobert Thingum

9711317

$endgroup$

share|cite|improve this answer

edited Mar 19 at 16:27

answered Mar 19 at 16:10

Robert ThingumRobert Thingum

9711317

answered Mar 19 at 16:10

Robert ThingumRobert Thingum

9711317

answered Mar 19 at 16:10

Robert ThingumRobert Thingum

9711317

1

$begingroup$

Let's suppose that $f$ is injective.

Let $x_1 neq x_2 in X$. Consider their images $f(x_1)$ and $f(x_2)$. They are different because $f$ is injective. Because $Y$ is Hausdorff, there exists two open subsets $U_1$ and $U_2$ of $Y$, with $f(x_1) in U_1$ and $f(x_2) in U_2$, and $U_1 cap U_2 = emptyset$.

Now $f^{-1}(U_1)$ and $f^{-1}(U_2)$ are open in $X$ by definition of open sets on $X$. They contain respectively $x_1$ and $x_2$. And they don't intersect, otherwise $U_1$ et $U_2$ would intersect.

This shows that $X$ is Hausdorff.

share|cite|improve this answer

edited Mar 19 at 16:10

answered Mar 19 at 16:07

TheSilverDoeTheSilverDoe

5,433216

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What if $f(x_{1})=f(x_{2})$?
  $endgroup$
  – Robert Thingum
  Mar 19 at 16:08
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for the precision. Indeed the property does not hold in general. I add that I suppose that $f$ is injective.
  $endgroup$
  – TheSilverDoe
  Mar 19 at 16:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  f:X-> Y is a "given mapping". What is the definition of "mapping" in topology?
  $endgroup$
  – user247327
  Mar 19 at 16:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user247327 What do you mean ? The property the OP wants to show is not true in general. I just give a proof when you add the asumption that $f$ is injective, in which case the property holds.
  $endgroup$
  – TheSilverDoe
  Mar 19 at 16:18
  

  
  
  
  

add a comment | 

1

$begingroup$

Let's suppose that $f$ is injective.

Let $x_1 neq x_2 in X$. Consider their images $f(x_1)$ and $f(x_2)$. They are different because $f$ is injective. Because $Y$ is Hausdorff, there exists two open subsets $U_1$ and $U_2$ of $Y$, with $f(x_1) in U_1$ and $f(x_2) in U_2$, and $U_1 cap U_2 = emptyset$.

Now $f^{-1}(U_1)$ and $f^{-1}(U_2)$ are open in $X$ by definition of open sets on $X$. They contain respectively $x_1$ and $x_2$. And they don't intersect, otherwise $U_1$ et $U_2$ would intersect.

This shows that $X$ is Hausdorff.

share|cite|improve this answer

edited Mar 19 at 16:10

answered Mar 19 at 16:07

TheSilverDoeTheSilverDoe

5,433216

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What if $f(x_{1})=f(x_{2})$?
  $endgroup$
  – Robert Thingum
  Mar 19 at 16:08
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you for the precision. Indeed the property does not hold in general. I add that I suppose that $f$ is injective.
  $endgroup$
  – TheSilverDoe
  Mar 19 at 16:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  f:X-> Y is a "given mapping". What is the definition of "mapping" in topology?
  $endgroup$
  – user247327
  Mar 19 at 16:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @user247327 What do you mean ? The property the OP wants to show is not true in general. I just give a proof when you add the asumption that $f$ is injective, in which case the property holds.
  $endgroup$
  – TheSilverDoe
  Mar 19 at 16:18
  

  
  
  
  

add a comment | 

$begingroup$

Let's suppose that $f$ is injective.

Let $x_1 neq x_2 in X$. Consider their images $f(x_1)$ and $f(x_2)$. They are different because $f$ is injective. Because $Y$ is Hausdorff, there exists two open subsets $U_1$ and $U_2$ of $Y$, with $f(x_1) in U_1$ and $f(x_2) in U_2$, and $U_1 cap U_2 = emptyset$.

Now $f^{-1}(U_1)$ and $f^{-1}(U_2)$ are open in $X$ by definition of open sets on $X$. They contain respectively $x_1$ and $x_2$. And they don't intersect, otherwise $U_1$ et $U_2$ would intersect.

This shows that $X$ is Hausdorff.

share|cite|improve this answer

edited Mar 19 at 16:10

answered Mar 19 at 16:07

TheSilverDoeTheSilverDoe

5,433216

$endgroup$

share|cite|improve this answer

edited Mar 19 at 16:10

answered Mar 19 at 16:07

TheSilverDoeTheSilverDoe

5,433216

answered Mar 19 at 16:07

TheSilverDoeTheSilverDoe

5,433216

answered Mar 19 at 16:07

TheSilverDoeTheSilverDoe

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This is not true in general. Suppose $f(x)=f(y)$, then for all $Usubset X$ open with $xin U$ there exists a $Vsubset Y$ open such that $U=f^{-1}(V)$. Note that $xin U$ if and only if $f(x)=f(y)in V$ which implies that $yin U$. So you need that $f$ is injective. If that holds you can follow the answer of TheSilverDoe.

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answered Mar 19 at 16:09

Floris ClaassensFloris Claassens

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$begingroup$

This is not true in general. Suppose $f(x)=f(y)$, then for all $Usubset X$ open with $xin U$ there exists a $Vsubset Y$ open such that $U=f^{-1}(V)$. Note that $xin U$ if and only if $f(x)=f(y)in V$ which implies that $yin U$. So you need that $f$ is injective. If that holds you can follow the answer of TheSilverDoe.

share|cite|improve this answer

answered Mar 19 at 16:09

Floris ClaassensFloris Claassens

1,33229

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add a comment | 

$begingroup$

This is not true in general. Suppose $f(x)=f(y)$, then for all $Usubset X$ open with $xin U$ there exists a $Vsubset Y$ open such that $U=f^{-1}(V)$. Note that $xin U$ if and only if $f(x)=f(y)in V$ which implies that $yin U$. So you need that $f$ is injective. If that holds you can follow the answer of TheSilverDoe.

share|cite|improve this answer

answered Mar 19 at 16:09

Floris ClaassensFloris Claassens

1,33229

$endgroup$

share|cite|improve this answer

answered Mar 19 at 16:09

Floris ClaassensFloris Claassens

1,33229

answered Mar 19 at 16:09

Floris ClaassensFloris Claassens

1,33229

answered Mar 19 at 16:09

Floris ClaassensFloris Claassens

1,33229