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Question

If

K1 and K2K_1\ and\ K_2

are disjoint nonempty compact sets, show that there exist

kiKik _ { i } \in K _ { i }

such that

0<k1k2=inf{x1x2:xiKi}.0 < \left| k _ { 1 } - k _ { 2 } \right| = \inf \left\{ \left| x _ { 1 } - x _ { 2 } \right| : x _ { i} \in K _ { i } \right\}.

Solution

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Let K1K_1 and K2K_2 be disjoint nonempty compact sets.

KiK_i is compact. Let (xn)(x_n) be a sequence such that xnKx_n\in K for every nNn\in \mathbb{N}. Because of Bolzano-Weierstrass Theorem 3.4.8. we know that there exists a subsequence (xnk)(x_{n_k}) and since KK is closed it follows that the limit of (xnk)(x_{n_k}) is in KK.

In other words, every sequence in KK has a subsequence that converges to a point that is in KK. Now, we know that KiK_i is bounded wich means that it has an infimum and supremum.

Let x1K1x_1\in K_1 and x2K2x_2\in K_2 and lets observe their difference:

x1x2 converges to K1K2|x_1-x_2| \text{ converges to } |K_1-K_2|

As K1K_1 and K2K_2 are bounded it follows that K1K2|K_1-K_2| is also bounded a therefore has infimum. We have:

k1k2=inf{x1x2 ; xiKi}\boldsymbol{ |k_1-k_2|=\inf \{|x_1-x_2|\ ; \ x_i \in K_i\}}

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