Definition of Hausdorff space and please check my proof

Question

I'm not sure about Hausdorff space property.

It's all point $x,y\in X$ with $x\neq y$ there exist open sets U containing $x$ and V or just some point $x,y$

Here this is my problem that I want to prove

Let X be a Hausdorff topological space, and Y be a subset of X. Prove that the subspace topology on Y is Hausdorff

my proof if it's all point $x,y$ in $X$

Let $X$ be Hausdorff space there exist open set $A$ and $B$ such that $x\in A,y\in B$ and $A\cap B\ =\varnothing $

Now we consider subspace $Y$

Since subspace $Y$ is subset of $X$ by definition of subspace

Then It's must exist

$i\in C,j\in D$ such that $i \neq j $that is $C\cap D=\varnothing$

Therefore subspace $Y$ is Hausdorff space too

Answer 1

The definition is: a topological space $X$ is a Hausdorff space if, for every $x,y\in X$ with $x\neq y$, there are open sets $U_x$ and $U_y$ such that:

• $x\in U_x$;
• $y\in U_y$;
• $U_x\cap U_y=\emptyset$.

So, if $Y\subset X$ and $x$ and $y$ are distinct elements of $Y$, you know, since you are assuming that $X$ is Hausdorff, that there are open sets $U_x$ and $U_y$ such that $x\in U_x$, $y\in U_y$, and $U_x\cap U_y=\emptyset$. Now, let $V_x=U_x\cap Y$ and let $V_y=U_y\cap Y$. Then

• $x\in V_x$;
• $y\in V_y$;
• $V_x\cap V_y=\emptyset$.

Therefore, $Y$ is Hausdorff.