I was asked to verify the truth of the statement $\text{int}(A_1 \cup A_2)=\text{int}A_1 \cup \text{int}A_2$, where the sets $A_i$ are subsets in a topological space $(X, \tau)$. I have already seen that the statement is in general not true, but I'd like to know what failed in my approach. My reasoning was the following.
Let $B$ denote the set of all open sets contained in $A_1 \cup A_2$. Let $B_\alpha$ denote a member of $B$. Then the interior of the union of said sets is the union of all the sets of $B$. Now let $C$ and $D$ be, respectivelly, the sets of all open sets contained in $A_1$ and $A_2$. Then, $B= C \cup D$. Then, $$\text{int}(A_1 \cup A_2)=\cup_\alpha B_\alpha=\cup_{\gamma,\beta} (C_\gamma \cup D_\beta)=(\cup_\gamma C_\gamma) \cup (\cup_\beta D_\beta)=\text{int}A_1 \cup \text{int}A_2$$
I am not able to see where I went wrong. Any help would be appreciated
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asked Dec 11, 2022 at 21:46
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$\begingroup$ The problem in this reasoning is that it's true just when $A_1$ and $A_2$ already is open. $\endgroup$– KempaCommented Dec 11, 2022 at 21:52
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$\begingroup$ Let $X=[0,1]$, $A_1=\mathbb{Q}\cap X$ and $A_2=A_1^c$, it's clear that int$(A_1\cup A_2)=X$, but int$(A_1)\cup$int$(A_2)=\emptyset$. $\endgroup$– KempaCommented Dec 11, 2022 at 21:55
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$\begingroup$ There are actually at least two separate conceptual errors here. To make matters worse, the "$B_\alpha,C_\gamma,D_\beta$" notation is confusing, making it hard to disentangle the two errors. Rather than try to correct everything at once, I'll just suggest two ideas. (1) Along with the set-theoretic formalism, keep in mind a simple example that can be visualised, such as two closed squares in the plane, joined by a common side. (2) The statement "$B=C\cup D$" isn't actually saying what you want it to say. It says that every open set contained in $A_1\cup A_2$ is contained in $A_1$ or in $A_2.$ $\endgroup$– Calum GilhooleyCommented Dec 11, 2022 at 22:09
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$\begingroup$ @CalumGilhooley Why not an official answer? $\endgroup$– Paul FrostCommented Dec 11, 2022 at 23:03
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$\begingroup$ @PaulFrost Because I had made the mistake of thinking I could just take a quick look at Maths.SE before bedtime! I just didn't have time to write a full answer, which would necessarily be quite long. So please do feel free to expand my comment into an answer, if you'd like to. (Ditto to anybody else.) $\endgroup$– Calum GilhooleyCommented Dec 11, 2022 at 23:43
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1 Answer
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You must recall that the interior of a set $X$ is union of all open sets in $X$, so when you assume $B=C\cup D$, you're saying that $\operatorname{int}(A_{1}\cup A_{2})=\operatorname{int}A_{1}\cup\operatorname{int}A_{2}$, which is wrong.
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$\begingroup$ But is it not true that the set of all open sets contained in $A_1$ joined with the set of all open sets contained in $A_2$ is the set of all open sets contained in their union? Are any of them missing? $\endgroup$ Commented Dec 11, 2022 at 22:00
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1$\begingroup$ Wait I just realized that I'm probably excluding the open sets that are not contained in neither but are contained in their union, so the correct statement would be $B \supset C \cup D$? $\endgroup$ Commented Dec 11, 2022 at 22:02
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1$\begingroup$ Exactly, $(0.5,1.5)\in [0,1]\cup(1,2)$ but $(0.5,1.5)\notin [0,1]$ and $(0.5,1.5)\notin(1,2)$ $\endgroup$ Commented Dec 11, 2022 at 22:09
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