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I know that $\operatorname{Int}(A) \cup \operatorname{Int}(B) \subset \operatorname{Int}(A \cup B)$, but that the other direction does not hold, so can anybody please tell me whats wrong with the following proof?

Suppose $x \in \operatorname{Int}(A \cup B) \Rightarrow \exists \epsilon > 0 : K(x, \epsilon) \subset A \cup B$. Assume without loss of generatlity that $K(x,\epsilon) \subset A-B \Rightarrow x \in \operatorname{Int}(A) \Rightarrow x \in \operatorname{Int}(A) \cup \operatorname{Int}(B)$.

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    $\begingroup$ How can you "assume without loss of generality" that $K(x,\epsilon) \subset A - B$ (I'm assuming $K(x, \epsilon)$ means a ball of radius $\epsilon$ or something similar)? $\endgroup$ – Pedro M. May 29 '15 at 12:57
  • $\begingroup$ I'm not sure whether you can acutally do that, but I thought that there would basically be three cases either: $K(x,\epsilon) \subset A-B, K(x,\epsilon) \subset B-A, K(x,\epsilon) \subset A \cap B.$ All of which would result in the same conclusion. $\endgroup$ – eager2learn May 29 '15 at 13:00
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    $\begingroup$ This is not the case, as many answers are pointing out. For instance, if $A = [0,1]$ and $B = [1,2]$, then $K = (1-\epsilon, 1+\epsilon)$ is contained in $A \cup B$ but does not fit in any of these three cases. $\endgroup$ – Pedro M. May 29 '15 at 13:02
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    $\begingroup$ @eager2learn Consider a different case, but that should shed light: $A=\{0,1,2\}$, $B=\{2,3,4\}$. Then $C=\{1,3\}$ is a subset of $A\cup B$, but $C$ is not contained in any of $A\setminus B$, $B\setminus A$ or $A\cap B$. It holds for elements that if $x\in A\cup B$ than one of the following is true: $x\in A\setminus B$, $x\in B\setminus A$ or $x\in A\cap B$. $\endgroup$ – egreg May 29 '15 at 13:07
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    $\begingroup$ You have also $A\cup B\subseteq A\cup B$. Is it correct to conclude that $A\cup B\subseteq A-B\vee A\cup B\subseteq A\cap B\vee A\cup B\subseteq B-A$? I don't think so. $\endgroup$ – drhab May 29 '15 at 13:08
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Because it's false, in general, that $\operatorname{Int}(A\cup B)\subseteq \operatorname{Int}(A)\cup \operatorname{Int}(B)$: take $A=(0,1]$ and $B=[1,2)$. Then $\operatorname{Int}(A)=(0,1)$, $\operatorname{Int}(B)=(1,2)$, but $\operatorname{Int}(A\cup B)=(0,2)$.

You can surely assume without loss of generality that $x\in A$, but not that the neighborhood $K(x,\epsilon)$ is contained is $A$. Try picturing what happens in the above example: $K(1,\epsilon)$ contains elements that are greater than $1$ (and belong to $B$) as well as elements that are less than $1$ (and belong to $A$). Since $A\cap B=\{1\}$, no neighborhood of $1$ can be contained either in $A$ or in $B$.


Consider a different case, but that should shed light over your difficulty: $X=\{0,1,2\}$, $Y=\{2,3,4\}$. Then $Z=\{1,3\}$ is a subset of $X\cup Y$, but $Z$ is not contained in any of $X\setminus Y$, $Y\setminus X$ or $X\cap Y$. It holds for elements that if $x\in X\cup Y$ than one of the following is true: $x\in X\setminus Y$, $x\in Y\setminus X$ or $x\in X\cap Y$.

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If $X\subseteq A\cup B$, it need not be the case that $X\subseteq A$ or $X\subseteq B$.

As for a specific case, in the real numbers both the rationals and their complement have empty interior. What is the interior of the union?

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  • $\begingroup$ Is there anything I can do to improve this answer? $\endgroup$ – Asaf Karagila May 29 '15 at 13:29
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    $\begingroup$ No thanks, I liked your answer, but egreg has updated his answer and I thought it was more helpful. Hope you dont mind that I accepted his now. $\endgroup$ – eager2learn May 29 '15 at 13:31
  • $\begingroup$ The comment wasn't aimed at you for unaccepting, it was aimed at whoever downvoted. As far as unaccepting goes, egreg wrote an excellent answer that's worth accepting. So that's fine. $\endgroup$ – Asaf Karagila May 29 '15 at 13:33
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$K(x,\epsilon) \subset A \cup B$ does not implies $K(x,\epsilon) \subset A \Delta B$. It is not true when $K(x,\epsilon) \cap (A \cap B) \neq \phi$. For example, consider $A = (0,2)$ and $B = (1,3)$ and $x = 1.5$ and $\epsilon = 0.25$.

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You assume without loss of generality that any point $x$ that lies in $A \cup B$ necessarily lies strictly in $A$ or strictly in $B$. If your point lies in the intersection $A \cap B$ your proof doesn't work. You have however proven the case where $A$ and $B$ are disjoint.

An extension of your proof by saying that the ball $K(x, \varepsilon)$ lies entirely in the intersection will not work. Look for instance at the following example: Suppose we endow $\mathbb{R}$ with the Euclidean topology, then we can look at the intervals $[0,1]$ and $[1,2]$. it is clear that $$1 \in Int([0,2])$$ But we have $$ 1 \notin Int([0,1]) \qquad 1 \notin Int([1,2]) $$ Thus providing a counterexample.

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$\operatorname{Int}([1,2]) \cup \operatorname{Int}([2,3]) =(1,2) \cup (2,3)$

$\operatorname{Int}([1,2] \cup [2,3]) = (1,3)$

$K(2, 1/10) \in \operatorname{Int}([1,2] \cup [2,3])$

There is no $\epsilon$ such that $K(2, \epsilon) \in [1,2]$ and $K(2, \epsilon) \in [2,3]$

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