A subspace of a Hausdorff space is Hausdorff

• September 24th 2011, 11:35 AM
dwsmith
A subspace of a Hausdorff space is Hausdorff
A subspace of a Hausdorff space is Hausdorff.

Let X be Hausdorff and $Y\subset X$

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

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

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

Thus, the subspace is Hausdorff.

Correct?
• September 24th 2011, 11:55 AM
Drexel28
Re: A subspace of a Hausdorff space is Hausdorff
Quote:

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

Let X be Hausdorff and $Y\subset X$

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

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

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

Thus, the subspace is Hausdorff.

Correct?

This isn't quite correct. The fact that $Y$ is a subspace of $X$ says that given any open subset $U\subseteq Y$ one has that $U={\color{red}{Y}}\cap V$ for some open set [etx]V[/tex] in $X$. want to take this and try again?
• September 24th 2011, 12:00 PM
Plato
Re: A subspace of a Hausdorff space is Hausdorff
Quote:

Originally Posted by dwsmith
A subspace of a Hausdorff space is Hausdorff.
Let X be Hausdorff and $Y\subset X$
Since Y is a subspace, $Y=X\cap C$ where C is some open set in X.

Suppose that $a\in Y~,~b\in Y~\&~a\ne b$. Then $\{a,b\}\subset X$.
But that means there are two disjoint open sets in $X$ such that $a\in\mathcal{O}_a~\&~b\in\mathcal{O}_b$ WHY?
What can you say about $Y\cap\mathcal{O}_a~\&~Y\cap\mathcal{O}_b~?$
• September 24th 2011, 12:06 PM
dwsmith
Re: A subspace of a Hausdorff space is Hausdorff
Quote:

Originally Posted by Drexel28
This isn't quite correct. The fact that $Y$ is a subspace of $X$ says that given any open subset $U\subseteq Y$ one has that $U={\color{red}{Y}}\cap V$ for some open set [etx]V[/tex] in $X$. 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 $x_1,x_2\in U$. Since U is open, disjoint sets $V_1,V_2$ can be constructed s.t. $x_1\in V_1 \ \text{and} \ x_2\in V_2$ and $V_1\cap V_2=\O$

Does this do it then?
• September 24th 2011, 12:25 PM
Drexel28
Re: A subspace of a Hausdorff space is Hausdorff
Quote:

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

Does this do it then?

Once again, not quite. You need to start with two arbitrary points $x_1,x_2\in Y$ and construct two disjoint neighborhoods of them. You know you can do this in $X$, in other words there exists disjiont open sets $V_1,V_2$ in $X$ with $x_i\in V_i$, now how can you project these down to open subsets in $Y$ which are disjoint and contain $x_i$?
• September 24th 2011, 12:35 PM
dwsmith
Re: A subspace of a Hausdorff space is Hausdorff
Quote:

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

Let $x_1,x_2\in Y$; $V_1,V_2\in X$, and $V_1\cap V_2=\O$ where $V_1,V_2$ are open in X.
$x_1\in V_1 \ \text{and} \ x_2\in V_2$

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

$U_1\cap U_2=\O$

Correct?
• September 24th 2011, 01:40 PM
Drexel28
Re: A subspace of a Hausdorff space is Hausdorff
Quote:

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

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

Correct?

Yes.