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I want to understand the highlighted part of the image below. Can someone help me in understanding this?

P.S. I am aware of this question. But I don't know compactness as of now. So please don't close or downvote this question saying that the answer already exists.

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Ryukendo Dey
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    The smallest closed sets in $X$ which intersection with $A$ is $C_i$ are $\overline{C_i}$. While our assumption implies $\overline{C_1}\cap C_2 = C_1\cap \overline{C_2} = \emptyset$, it can still happen that $\overline{C_1}\cap \overline{C_2}\neq\emptyset$. – Jakobian Jul 01 '23 at 15:52

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$D_1$ and $D_2$ can happen to always have intersection: you can imagine $X$ to be $A\cup\{\infty\}$ for a new point $\infty$, and set that open sets in $X$ to be the open sets in $A$ plus the entire $X$. Then every nonempty closed set in $X$ contains $\infty$.

This is exactly the same construction as used in their previous proof that every topological space is a subspace of a normal space, by the way.

  • why do you say that the open sets in $X$ are the open sets in $A$ plus the entire $X$ – Ryukendo Dey Jul 01 '23 at 16:34
  • I am saying something slightly different. Let's say $A$ is a given topological space with the set of open sets $\mathcal A$. I propose to construct $X$ as $X:=A\cup{\infty}$ with the set of open sets $\mathcal X:=\mathcal A\cup{X}$. Now it is easy to show that $(A,\mathcal A)$ is a subspace of $(X,\mathcal X)$ but every non empty closed set in the latter contains the newly-added element $\infty$. To emphasise: I don't start with $A$ and $X$ given; I start with $A$ alone, and then construct $X$ in a particular way. –  Jul 01 '23 at 22:58
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I am posting this as an answer because of the length constraint in the comments.

So, I found out the deal here. Basically, I came across this idea while trying to prove the following:

Every closed subspace of a normal space is normal.

To prove this, observe that if $A \subseteq X$ where $(X, \mathcal{T})$ is a normal topological space and if $C, D \subseteq A$ are closed and disjoint in $A$ then they are also closed in $X$. So we can separate them by $U$ and $V$ in $X$ (as $X$ is normal). But then $U\cap A$ and $V \cap A$ are the separating and disjoint open sets in $A$.

Now coming back to my question. In my question $A$ need not be closed. So, $C$ and $D$ need not be closed in $X$. But normality is for separating closed sets from themselves. So, what are we even trying to separate?

Ryukendo Dey
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