Let $X$ be a Hausdorff compact space; let $\{A_n\}$ be a countable collection of closed set of $X$. Show that if each set $A_n$ has empty interior in $X$, then the union $\bigcup A_n$ has empty interior in $X$.

My solution:

Suppose that, $U=$ interior of $\bigcup A_n$ is nonempty. Then any open subset of $U$ can not have isolated point, since then all interior of $A_n$ can not be empty. (If $U$ have isolated point, then for some $x\in X$ with $\{x\}\subset U$ is an open set, then it must be a subset of some $A_n$, since $\{x\}$ do not intersects with $X-A_n$).

Let, $x_1\in A_1$. Choose a point $x_2\in U$ that do not contain in $A_1$. Now choose disjoint sets $W_1$ and $W_2$ about $x_1$ and $x_2$. Now, take $V_1=W_2\cap U\cap A^c_1$. Then, $x_1$ do not contain in $\bar V_1$.

Again do this: There exist an $k_1$ such that $x_2\in A_{k_1}$. Choose a point $x_3\in V_1$ that do not contain in $A_{k_1}$. Now choose disjoint sets $W_3$ and $W_4$ about $x_2$ and $x_3$. Now, take $V_2=W_4\cap U\cap A^c_{k_1}$. Then, $x_2$ do not contain in $\bar V_2$.

In this way we can find a collection of closed sets $\{\bar V_n\}$ such that $$\bar V_1\supset \bar V_2\supset\bar V_3\supset\dots$$

Note that, since $X$ is compact, $\bigcap \bar V_n$ is nonempty. Let, $x\in \bigcap \bar V_n$.

Interior of $A_n$ is empty and $x\notin A_n$ for all $n$ leads to a contradiction.

Is this solution ok? Any help appreciated.

Sorry that I published same question twice. I marked the other one duplicate.

  • $\begingroup$ For future reference, you can edit the question, instead of posting a new one. $\endgroup$ – Asaf Karagila Nov 5 '17 at 8:48
  • $\begingroup$ Thanks @AsafKaragila , but need to know if my solution ok or not. The previous one is wrong (which I realized after a long time). I will edit next time, but I was not sure anyone would see the edited version, since it was posted a at least 8 hrs ago. $\endgroup$ – topology_001 Nov 5 '17 at 11:16
  • 1
    $\begingroup$ (1) The front page is generally renewed every couple of hours anyway. (2) Editing bumps your question to the front page. (3) Most users browse tags rather than the front page directly (or at least in addition), where your question can be seen for longer, and by more people. (4) There are ways to call attention to your question anyway, reposting it and closing as a duplicate is not one of them, and can be grounds for suspension if done repeatedly. $\endgroup$ – Asaf Karagila Nov 5 '17 at 11:20
  • $\begingroup$ See here: math.stackexchange.com/questions/1296634/… $\endgroup$ – Math1000 Nov 7 '17 at 8:44

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