'Fake' identity regarding the closure in the subspace topology

Question

I have the following argument which I encountered, and can't seem to find why it's not true:

Let $X$ be a topological space and let $A$ and $B$ be nonempty subsets of $X$. Then $\overline{A\cap B}^A=\overline{A\cap B}^X\cap A$.

"Proof"

$\overline{A\cap B}^A =\cap \{ F\subset A: \; F\supseteq A\cap B, F \text{ is closed in } A \}$.

Since $F$ is closed in $A$ if and only if $F=F'\cap A$ where $F$ is closed in $X$. Hence

$\overline{A\cap B}^A= \cap \{ F'\cap A: \; F'\supseteq A\cap B, F' \text{ is closed }\}= \Big( \cap \{ F': \; F'\supseteq A\cap B, F' \text{ is closed }\}\Big) \cap A $.

And using an equivalent characterization of the closure,

$ \cap \{ F': \; F'\supseteq A\cap B, F' \text{ is closed }\}=\overline{A\cap B} $.

And finally we conclude

$ \overline{A\cap B}^A=\overline{A\cap B}^X\cap A $.

I can't seem to find where this argument falters, but this thread makes me think that it's wrong, and would appreciate any pointers.

What makes you think it's wrong? In general $\overline{B}^A=\overline{B}^X\cap A$. — Chrystomath, Jul 15 '20 at 12:19
@Chrystomath I think your "in general remark" is wrong. When $X=\mathbb{R},$ $B=\mathbb{Q}$ and $A\cap B=\emptyset$, this is not true. Which is why I think my identity is wrong. — Keen-ameteur, Jul 15 '20 at 12:21
@Chrystomath $A=\mathbb{R}\setminus \mathbb{Q}$ for example. — Keen-ameteur, Jul 15 '20 at 12:23
See https://math.stackexchange.com/questions/268943/closure-of-a-subset-of-a-subspace-of-a-topological-space#268945 — Chrystomath, Jul 15 '20 at 12:23
Then $\overline{\varnothing}^A=\varnothing=\overline{\varnothing}^X\cap A$ — Chrystomath, Jul 15 '20 at 12:24
@Chrystomath But $B\neq \emptyset$ and $\overline{B}^X=X$ in my case. — Keen-ameteur, Jul 15 '20 at 12:28
If $A=\mathbb{R}\setminus\mathbb{Q}$ and $B=\mathbb{Q}$ then $A\cap B=\varnothing$ not $B$. — Chrystomath, Jul 15 '20 at 12:29
That thread is different. It says $\overline{A\cap B}^A=\overline{B}^X\cap A$. — Chrystomath, Jul 15 '20 at 12:36
Read carefully the difference between the question you linked and yours. In your notation, it is claiming that $\overline{A \cap B}^A = \overline{B}^X \cap A$ which is false. You are claiming instead that $\overline{A \cap B}^A = \overline{A \cap B}^X \cap A$ which is true. — Nate Eldredge, Jul 15 '20 at 12:37
$\overline B^A=\overline B^X\cap A$ when $B\subseteq A.$ But $\overline B^A$ is meaningless when $B\not\subseteq A.$ — DanielWainfleet, Jul 17 '20 at 02:45

score 1 · Accepted Answer · answered Jul 15 '20 at 12:24

1

I am not sure whether this answers your question but in general if $D\subseteq A$ then $\overline D^A=A\cap\overline D$.

Proof:

Observe that $A\cap\overline D$ is closed in $A$ with $D\subseteq A\cap\overline D$ telling us that $\overline D^A\subseteq A\cap\overline D$.

Conversely $\overline D^A$ is closed in $A$ so that $\overline D^A=A\cap F$ for some closed set $F$.

Then from $D\subseteq\overline D^A$ it follows that $D\subseteq F$ and consequently $\overline D\subseteq F$.

Then $A\cap\overline D\subseteq A\cap F=\overline D^A$.

Applying this on $D:=A\cap B$ we find the statement in your question.

answered Jul 15 '20 at 12:24

drhab

151,093

So you think the statement is true? – Keen-ameteur Jul 15 '20 at 12:25
1

Indeed. The statement is true. – drhab Jul 15 '20 at 12:26
Just one last thing, do you think the reasoning I wrote is generally sound? – Keen-ameteur Jul 15 '20 at 12:34
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If I read it correctly then you state:$$\overline{A\cap B}^{A}=\bigcap\left{ A\cap F'\mid F'\text{ closed and }A\cap B\subseteq F'\right} =A\cap\bigcap\left{ F'\mid F'\text{ closed and }A\cap B\subseteq F'\right} =A\cap\overline{A\cap B}$$This is sound. – drhab Jul 15 '20 at 12:40

'Fake' identity regarding the closure in the subspace topology

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