# Why is my textbook using a rather long proof involving sequences?

Claim: If $K_1$ and $K_2$ are disjoint nonempty compact sets, show that $\exists k_i\in K_i$ such that $0<\vert k_1-k_2\vert=\inf\{\vert x_1-x_2\vert : x_1\in K_1 \quad\land\quad x_2\in K_2\}$.

Can't we simply say that if $0=\vert y_1-y_2\vert=\inf\{\vert x_1-x_2\vert : x_1\in K_1 \quad\land\quad x_2\in K_2\}$, then $y_1\in K_1\cap K_2$ - a contradiction?

here what the book provided:

• But how do you define $y_1$ and $y_2$? You did not justify that your infimum may be written as $|y_1-y_2|$ with $y_1\in K_1$ and $y_2\in K_2$. However, this is required by the claim. – Suzet Jul 22 '18 at 5:50
• No. You need to use compactness somehow. e.g., Consider $K_1 = \{0\}$ and $K_2 = \{1/n | n \in \mathbb{N}\}$. – Jair Taylor Jul 22 '18 at 5:50
• The non-trivial claim is that the $\inf$ is attained by some $k_1, k_2$, not that the $\inf$ is strictly positive – Rhys Steele Jul 22 '18 at 5:50
• I presume this is because the book has not proved one of (i) the product of two compact spaces is compact, (ii) the image of a compact set under a continuous map is compact, (iii) a compact subset of $\Bbb R$ is closed. – Lord Shark the Unknown Jul 22 '18 at 5:56
• inf is not min. You have no idea that any $|y_1 - y_2| = \inf$ exist. – fleablood Jul 22 '18 at 5:58

$A < B = C$ is a shorthand for $A < B$ and $B = C$.
You seem to think that its negation is $A \ge B = C$ (a shorthand for $A \ge B$ and $B = C$)
But that is not the case : its negation is $A \ge B$ or $B \neq C$.
So it is not valid to prove $A < B = C$ by assuming $A \ge B = C$ and getting a contradiction.
$K_1$ and $K_2$ are compact, so $K_1 \times K_2$ is compact by Tychonoff. The function $K_1 \times K_2 \to \Bbb R$ sending $(x_1, x_2)$ to $|x_1 - x_2|$ is continuous, so the range is compact, so it has a minimum which can be expressed as $|k_1 - k_2|$ for some $k_1 \in K_1$ and $k_2 \in K_2$.