# Why is the interior of an union not the union of the interiors

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

• The problem in this reasoning is that it's true just when $A_1$ and $A_2$ already is open. Commented Dec 11, 2022 at 21:52
• 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$(A_2)=\emptyset$. Commented Dec 11, 2022 at 21:55
• 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 $A_2.$ Commented Dec 11, 2022 at 22:09
• @CalumGilhooley Why not an official answer? Commented Dec 11, 2022 at 23:03
• @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.) Commented Dec 11, 2022 at 23:43

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.
• 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? Commented Dec 11, 2022 at 22:00
• 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 $B \supset C \cup D$? Commented Dec 11, 2022 at 22:02
• Exactly, $(0.5,1.5)\in [0,1]\cup(1,2)$ but $(0.5,1.5)\notin [0,1]$ and $(0.5,1.5)\notin(1,2)$ Commented Dec 11, 2022 at 22:09