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Let $A$ and $B$ be closed, bounded, disjunct subsets of $\mathbb{R}^p$

Now, this is not a metric, but define $\delta$ like this: $$ \delta = \inf V, $$ where $$ V = \{ \| a-b \| \mid a \in A \text{ and } b \in B \} $$

Show that $\delta$ is strictly greater than $0$.

Here's my reasoning: Since $A$ and $B$ are closed and bounded, $V$ will be closed and bounded. Because $V$ is closed and bounded, it will contain its infimum. Because $A$ and $B$ are disjunct, $0$ can't be part of $V$. Because $V$ only contains positive members of $\mathbb{R}$, $\delta$ cannot be $0$ and must therefore be strictly positive.

The part where I'm confused is the first sentence. Why exactly will $V$ be closed?

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Briefly to say, if $\inf V=0$, then there are sequences $\{a_i\}, \{b_i\}$ such that $\Vert a_i-b_i\Vert$ tend to zero. But, then, by the boundedness of $A,B$, we can find convergent subsequences of $\{a_{k_i}\}, \{b_{k_i}\}$ respectively, and these two subsequences will have the same limit, which will be in $A\cap B$ by the closedness of $A,B$, it contradicts the disjointness of them.

To your confusion of closedness of $V$, one way to show it is to use the sequential completeness, as in my answer.

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