# Thread: A subspace of a Hausdorff space is Hausdorff

1. ## 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?

2. ## Re: A subspace of a Hausdorff space is Hausdorff

Originally Posted by dwsmith
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?

3. ## Re: A subspace of a Hausdorff space is Hausdorff

Originally Posted by dwsmith
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 $\displaystyle Y\cap\mathcal{O}_a~\&~Y\cap\mathcal{O}_b~?$

4. ## Re: A subspace of a Hausdorff space is Hausdorff

Originally Posted by Drexel28
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?

5. ## Re: A subspace of a Hausdorff space is Hausdorff

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 $\displaystyle x_1,x_2\in U$. 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 $\displaystyle x_i$?

6. ## Re: A subspace of a Hausdorff space is Hausdorff

Originally Posted by Drexel28
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?

7. ## Re: A subspace of a Hausdorff space is Hausdorff

Originally Posted by dwsmith
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.