intersection of two compact subsets in a space may be non-compact and not closed

hbghlyj

topology HT 2009 Exercise 6.
(1) Let $X$ be a topological space and $A$ a subset of $X$. On $X \times\{0,1\}$ define the partition composed of the pairs $\{(a, 0),(a, 1)\}$ for $a∈A$, and of the singletons $\{(x, i)\}$ if $x∈X∖A$ and $i ∈\{0,1\}$.
Let $\mathcal{R}$ be the equivalence relation defined by this partition, let $Y$ be the quotient space $[X \times\{0,1\}] /\mathcal{R}$ and let $p: X \times\{0,1\}→Y$ be the quotient map.

(a) Prove that there exists a continuous map $f: Y→X$ such that $f \circ p(x, i)=x$ for every $x∈X$ and $i ∈\{0,1\}$.

(b) Prove that $Y$ is Hausdorff if and only if $X$ is Hausdorff and $A$ is a closed subset of $X$.

(2) Consider the above construction for $X=[0,1]$ and $A$ an arbitrary subset of $[0,1]$.
Prove that $Y$ is compact. Prove that $K=p(X \times\{0\})$ and $L=p(X \times\{1\})$ are compact, and that $K ∩L$ is homeomorphic to $A$.
We have thus shown that the intersection of two compact subsets in a space that is not Hausdorff may be non-compact and not closed.

hbghlyj

math.stackexchange.com/questions/1446755

Answer to part 1b (attempt)

"$\Rightarrow$": consider the function $p:X\times \{0,1\} \rightarrow (X\times \{0,1\})/ R$, and suppose $X$ isn't Hausdorff. Take $(x_1,0)\neq (x_2,0)$ and there exist separated open sets $U,V$ containing $p(x_1,0)$ and $p(x_2,0)$ respectively. Then $p^{-1}(U),p^{-1}(V)$ are open in $X\times \{0,1\}$ by continuity of $p$, are separated and contain respectively $(x_1,0)$ and $(x_2,0)$, contadicting that $X$ is Hausdorff.

Now, suppose $A$ isn't closed. Take a point $a$ on the boundary of $A$, and consider $(a,0),(a,1)$ in $X\times \{0,1\}$. $p(a,0)\neq  p(a,1)$. I will show these points cannot be separated. Intuitively, I need to use the fact that any open neighbourhood of $a$ contains a point $b\in A$, and that $p(b,0)=p(b,1)$, but I'm having difficulty articulating an argument.

"$\Leftarrow$": if $(x,i_x),(y,i_y)$ are distinct and separable in $X \times \{0,1\}$, then clearly so are $p(x,i_x)$ and $p(y,i_y)$ in $(X\times \{0,1\})/ R$. If $x\in X\backslash A$, then $(x,0)\neq(x,1)$. As $A$ is closed, $X\backslash A$ is open. Then just take the sets $X\backslash A\times\{0\}$ and $X\backslash A\times\{1\}$ as separating sets.

Please help me with the proof that $A$ is closed

hbghlyj

Maybe useful:
$X/{\sim}$ is Hausdorff if and only if $\sim$ is closed in $X \times X$

Czhang271828

hbghlyj 发表于 2023-2-16 23:57
math.stackexchange.com/questions/1446755

Please help me with the proof that $A$ is closed ...

Hint: $a\in (X,\tau)$ is separable, whenever
$$
\bigcap\{U\in \tau\mid a\in U\}=\{a\}.
$$

If $A$ is not closed, then there exists $b\in \overline A\setminus A$. Consider the topology of $Y$, we have
$$
\begin{align*}
&\bigcap\{U\in \tau _Y\mid (b,0)\in U\}\\
=\,&\bigcap\{U\in \tau _Y\mid (b,0)\in U,U\cap A\times \{0,1\}\neq \varnothing\}\\
=\,&\bigcap\{U\in \tau _Y\mid (b,0)\in U,(b,1)\in U,U\cap A\times \{0,1\}\neq \varnothing\}\\
=\,&\bigcap\{U\in \tau _Y\mid (b,1)\in U,U\cap A\times \{0,1\}\neq \varnothing\}\\
=\,&\bigcap\{U\in \tau _Y\mid (b,1)\in U\}.
\end{align*}
$$
As a result, $(b,0)$ and $(b,1)$ are......

我想, 以上解法不能更精炼了.