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?

Re: A subspace of a Hausdorff space is Hausdorff

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**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?

Re: A subspace of a Hausdorff space is Hausdorff

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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 $\displaystyle Y\cap\mathcal{O}_a~\&~Y\cap\mathcal{O}_b~?$

Re: A subspace of a Hausdorff space is Hausdorff

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**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?

Re: A subspace of a Hausdorff space is Hausdorff

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**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 $\displaystyle x_i$?

Re: A subspace of a Hausdorff space is Hausdorff

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**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?

Re: A subspace of a Hausdorff space is Hausdorff

Quote:

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.