Is closedness assumption is essential in following argument?

Question

$F$ is closed subset of $X$. $A$ is any subset of $X$. $X$ is a metric space. The topology on $X$ is induced by the metric.

Define $A_n= \{ x\in F^c\cap A\mid d(x,F)\ge \frac{1}{n}\}$

Then $\bigcup_{n=1}^{\infty} A_n= F^c\cap A$.

I had the above argument. To show the claim, is it necessary to have closedness assumption of $F$ ?

Or is there any counterexample such that above assertion fails if we relax closedness argument?

Any help will be appreciated.

Answer 1

You have missed $A$ in the definition of $A$. Assume that $A=X$. Then union of $A_n$'s is $\{x:d(x,F^{c}) >0\}$ and this is equal to $F^{c}$ iff $F$ is closed. For a counterexample you can take any set which is not closed!.

Answer 2

It is essential.

Suppose $x \in \bigcup_n A_n$. So $x \in A_m$ for some $m$. It follows that $x \in A$ and $d(x,F) > \frac{1}{m} >0$. So $x \notin F$ (as $x \in F$ implies $d(x,F)=0$), and so $x \in A \cap F^\complement$. This direction does not use the closedness of $F$.

Suppose $x \in F^\complement \cap A$. Then as $F$ is closed and $F^\complement$ is open, there is some $r>0$ such that $B(x,r) \subseteq F^\complement$, or $B(x,r) \cap F=\emptyset$. If we choose $m$ such that $\frac{1}{m}< r$, we see that $d(x,F) \ge r > \frac{1}{m}$ so that $x \in A_m$ and so $x \in \bigcup_n A_n$ as required. This inclusion does use the closedness of $F$ explicitly.

In the reals, take $A=\{0,1\}$ and $F=(0,1)$ which is not closed. Note that $d(x,F)=0$ for all $x \in A$.

Then for any $n$, $A_n = \emptyset$, so the equality fails rather miserably, as $A \cap F^\complement = A = \{0,1\}$.