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$\newcommand{\Int}{\operatorname{Int}}\newcommand{\Bdy}{\operatorname{Bdy}}$ If $A$ and $B$ are sets in a metric space, show that: (note that $\Int$ stands for interior of the set)

  1. $\Int (A) \cup \Int (B) \subset \Int (A \cup B)$.
  2. $(\overline{ A \cup B}) = (\overline A \cup \overline B )$. (note that $\overline A = \Int (A) \cup \Bdy(A)$ )

Now for the first (1) I see why its true for instance in $R$ we can have the intervals set $A=[a,b]$ and $B=[b,c]$ we have $A \cup B=[a,c]$ so $\Int(A \cup B)=(a,c)$ now $\Int(A)=(a,b)$ and $\Int(B)=(b,c)$ so we lose $b$ when we take union to form $\Int(A) \cup \Int(B)=(a,b) \cup (b,c)$.

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Related question (the poster asked about real line there): Show that $\operatorname{int}(A \cap B)= \operatorname{int}(A) \cap \operatorname{int}(B)$ – Martin Sleziak Oct 30 '12 at 6:21

For number two, I would recommend demonstrating it with Venn Diagrams. Have two overlapping circles labeled A and B. Via shading, show that the same space is shaded for each side of the equation.

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What happens if A and B are disjoint though? – drew Oct 30 '12 at 4:29
Honestly, I don't think it matters. Whatever is not in the union of A and B is the same as that which is in neither A nor B. – FrodoHackins Oct 30 '12 at 4:38

You have made a good start on (1). However, given two intervals $[a, b]$ and $[c, d]$ on $\mathbb{R}$ why should it be the case that $b = c$? Moreover, why do two sets on $\mathbb{R}$ even need to be closed intervals?

What if $A = \mathbb{N}$. Then we still have $A \subset \mathbb{R}$!

Thus you should go back to the definition of the interior!

Suppose $x \in \Int(A) \cup \Int(B)$. What does this mean? Once we know what it means, can we show why $x \in \Int(A \cup B)$? In this way, you will also avoid needing any of the special characteristics of $\mathbb{R}$!

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For your first question if $x \in Int(A) \cup Int(B)$ so x is in the open ball B(a;r)={y:d(a,y)<r}, and d(a,x)<r and the open ball is either in Int(A) or Int(B). Since Int(AUB) is an open set there exists a ball B(a;r) such that for the interior points in AUB they are in some ball B(a;r). – drew Oct 30 '12 at 4:48
Would you mind using latex? It's a little hard for me to read. – Alex Oct 30 '12 at 4:50

Recall that $\operatorname{Int}(A) \subset A$ for any set $A$ and that the interior is the largest such open set contained in $A$. Similarly, $\overline A$ is the smallest closed set containing $A$.

Then for $(a)$, $\operatorname{Int}(A) \cup \operatorname{Int}(B) \subseteq A \cup B$ and is an open set, so $\operatorname{Int}(A) \cup \operatorname{Int}(B) \subseteq \operatorname{Int}(A \cup B)$.

For $(b)$, $\overline{A} \cup \overline{B}$ is a closed set containing $A \cup B$, so $\overline{A \cup B} \subseteq \overline{A} \cup \overline{B}$. For the other side of the inclusion, since $\overline{A} \subseteq \overline{A \cup B}$ and $\overline{B} \subseteq \overline{A \cup B}$, we have $\overline{A} \cup \overline{B} \subseteq \overline{A \cup B}$.

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