Results 1 to 7 of 7

Thread: A subspace of a Hausdorff space is Hausdorff

  1. #1
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    10

    A subspace of a Hausdorff space is Hausdorff

    A subspace of a Hausdorff space is Hausdorff.

    Let X be Hausdorff and $\displaystyle Y\subset X$

    Since Y is a subspace, $\displaystyle Y=X\cap C$ where C is some open set in X.

    Let $\displaystyle x_1,x_2\in X\cap C$

    $\displaystyle X\cap C$ is the intersection of open sets so it is open. Therefore, $\displaystyle x_1\in U \ \text{and} \ x_2\in V$ where U and V are open sets in $\displaystyle X\cap C$. Again since $\displaystyle X\cap C$ is open, U and V can be constructed in such a manner that $\displaystyle U\cap V=\O$.

    Thus, the subspace is Hausdorff.

    Correct?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    22

    Re: A subspace of a Hausdorff space is Hausdorff

    Quote Originally Posted by dwsmith View Post
    A subspace of a Hausdorff space is Hausdorff.

    Let X be Hausdorff and $\displaystyle Y\subset X$

    Since Y is a subspace, $\displaystyle Y=X\cap C$ where C is some open set in X.

    Let $\displaystyle x_1,x_2\in X\cap C$

    $\displaystyle X\cap C$ is the intersection of open sets so it is open. Therefore, $\displaystyle x_1\in U \ \text{and} \ x_2\in V$ where U and V are open sets in $\displaystyle X\cap C$. Again since $\displaystyle X\cap C$ is open, U and V can be constructed in such a manner that $\displaystyle U\cap V=\O$.

    Thus, the subspace is Hausdorff.

    Correct?
    This isn't quite correct. The fact that $\displaystyle Y$ is a subspace of $\displaystyle X$ says that given any open subset $\displaystyle U\subseteq Y$ one has that $\displaystyle U={\color{red}{Y}}\cap V$ for some open set [etx]V[/tex] in $\displaystyle X$. want to take this and try again?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,776
    Thanks
    2823
    Awards
    1

    Re: A subspace of a Hausdorff space is Hausdorff

    Quote Originally Posted by dwsmith View Post
    A subspace of a Hausdorff space is Hausdorff.
    Let X be Hausdorff and $\displaystyle Y\subset X$
    Since Y is a subspace, $\displaystyle Y=X\cap C$ where C is some open set in X.
    Suppose that $\displaystyle a\in Y~,~b\in Y~\&~a\ne b$. Then $\displaystyle \{a,b\}\subset X$.
    But that means there are two disjoint open sets in $\displaystyle X$ such that $\displaystyle a\in\mathcal{O}_a~\&~b\in\mathcal{O}_b$ WHY?
    What can you say about $\displaystyle Y\cap\mathcal{O}_a~\&~Y\cap\mathcal{O}_b~?$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    10

    Re: A subspace of a Hausdorff space is Hausdorff

    Quote Originally Posted by Drexel28 View Post
    This isn't quite correct. The fact that $\displaystyle Y$ is a subspace of $\displaystyle X$ says that given any open subset $\displaystyle U\subseteq Y$ one has that $\displaystyle U={\color{red}{Y}}\cap V$ for some open set [etx]V[/tex] in $\displaystyle X$. want to take this and try again?
    From that, we know U is open in X since it is the intersection of two open sets in X. Let $\displaystyle x_1,x_2\in U$. Since U is open, disjoint sets $\displaystyle V_1,V_2$ can be constructed s.t. $\displaystyle x_1\in V_1 \ \text{and} \ x_2\in V_2$ and $\displaystyle V_1\cap V_2=\O$

    Does this do it then?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    22

    Re: A subspace of a Hausdorff space is Hausdorff

    Quote Originally Posted by dwsmith View Post
    From that, we know U is open in X since it is the intersection of two open sets in X. Let $\displaystyle x_1,x_2\in U$. Since U is open, disjoint sets $\displaystyle V_1,V_2$ can be constructed s.t. $\displaystyle x_1\in V_1 \ \text{and} \ x_2\in V_2$ and $\displaystyle V_1\cap V_2=\O$

    Does this do it then?
    Once again, not quite. You need to start with two arbitrary points $\displaystyle x_1,x_2\in Y$ and construct two disjoint neighborhoods of them. You know you can do this in $\displaystyle X$, in other words there exists disjiont open sets $\displaystyle V_1,V_2$ in $\displaystyle X$ with $\displaystyle x_i\in V_i$, now how can you project these down to open subsets in $\displaystyle Y$ which are disjoint and contain $\displaystyle x_i$?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    10

    Re: A subspace of a Hausdorff space is Hausdorff

    Quote Originally Posted by Drexel28 View Post
    Once again, not quite. You need to start with two arbitrary points $\displaystyle x_1,x_2\in Y$ and construct two disjoint neighborhoods of them. You know you can do this in $\displaystyle X$, in other words there exists disjiont open sets $\displaystyle V_1,V_2$ in $\displaystyle X$ with $\displaystyle x_i\in V_i$, now how can you project these down to open subsets in $\displaystyle Y$ which are disjoint and contain $\displaystyle x_i$?
    Let $\displaystyle x_1,x_2\in Y$; $\displaystyle V_1,V_2\in X$, and $\displaystyle V_1\cap V_2=\O$ where $\displaystyle V_1,V_2$ are open in X.
    $\displaystyle x_1\in V_1 \ \text{and} \ x_2\in V_2$


    $\displaystyle Y\cap V_1=U_1 \ \text{and} \ Y\cap V_2=U_2$.

    $\displaystyle U_1\cap U_2=\O$

    Correct?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    22

    Re: A subspace of a Hausdorff space is Hausdorff

    Quote Originally Posted by dwsmith View Post
    Let $\displaystyle x_1,x_2\in Y$; $\displaystyle V_1,V_2\in X$, and $\displaystyle V_1\cap V_2=\O$ where $\displaystyle V_1,V_2$ are open in X.
    $\displaystyle x_1\in V_1 \ \text{and} \ x_2\in V_2$


    $\displaystyle Y\cap V_1=U_1 \ \text{and} \ Y\cap V_2=U_2$.
    Yes.
    $\displaystyle U_1\cap U_2=\O$

    Correct?
    Yes.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Hausdorff space
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: Sep 19th 2011, 04:16 PM
  2. Hausdorff and Compact Space
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Jan 29th 2010, 03:13 AM
  3. Quotient map from Hausdorff to non-Hausdorff space
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Nov 28th 2009, 05:51 PM
  4. Hausdorff space, structure
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Mar 22nd 2009, 06:12 PM
  5. Compactness in a Hausdorff space
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: Feb 25th 2009, 08:59 AM

Search tags for this page

Click on a term to search for related topics.

/mathhelpforum @mathhelpforum