let $X$ be a topological space and let A be a topological subspace of $X $,
Is the following statement is true /false ?
" if $A $ is Hausdorff then $\bar A$ is also hausdorff "
My attempt :yes by the theorem every subspace of a hausdorff space is hausdorff
Take $X = \mathbb{R}$ and $A = \mathbb{Q}$
Consider the set $X = \left\lbrace a, b, c \right\rbrace$ with the topology $\mathscr{T} = \left\lbrace \emptyset, \left\lbrace a \right\rbrace, \left\lbrace b \right\rbrace, \left\lbrace a, b \right\rbrace, X \right\rbrace$. Now, let $A = \left\lbrace a, b \right\rbrace$. Clearly, $A$ is Hausdorff since $a \in \left\lbrace a \right\rbrace \in \mathscr{T}$ and $b \in \left\lbrace b \right\rbrace \in \mathscr{T}$ and $\left\lbrace a \right\rbrace \cap \left\lbrace b \right\rbrace = \emptyset$.
Now, the only open set containing $c$ is $X$ itself and hence has a non - empty intersection with $A$. Therefore, $\bar{A} = X$.
Therefore, in our example, $A$ is Hausdorff but $\bar{A} = X$ is not!
The line with two origins is such an example.
To make it clear, consider the real line and delete $0$. Now, in place of $0$, we put $p$ and $q$. We'll call this space $\Bbb R^\star$. Take $A=\Bbb R^\star\setminus\{p,q\}$; since any open set containing $p$ will intersect $A$, we have $p\in\overline{A}$. Similarly, $q\in\overline{A}$, but there is not an open set $U$ containing $p$ and an open set $V$ containing $q$ such that $U\cap V=\varnothing$, so $\overline{A}$ is not Hausdorff.
Consider the unit circle $S^1$, i.e. $\{e^{i\theta}|0\leq \theta< 2\pi\}$. We put on it the topology whose open sets are $\emptyset$, $S^1$, and any open sets that does not contain $1=e^{i\cdot 0}=e^{i\cdot 2\pi}$, i.e. $U\neq \emptyset, S^1$ is open iff $U=\{e^{i\theta}\,|\, \theta\in S\}$ where $S$ is an open set in $(0, 2\pi)$ using the usual topology on ${\mathbb R}$. Now let $A=S^1-\{1\}$, it is Hausdorff but its closure $\overline{A}=S^1$ is not.
