Let $A$ be a nonempty compact subset of $\mathbb{R}$ and $c \in \mathbb{R}$.
Prove that there exists a point $a$ in $A$ such that $| c-a | =\inf \{| c-x |: x \in A \}$?
$\begingroup$
$\endgroup$
1
-
8$\begingroup$ Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, many users find the use of the imperative ("Prove", "Show", etc) to be rude when asking for help. Please consider rewriting your post. $\endgroup$– Arturo MagidinCommented May 13, 2012 at 3:43
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
Hint. For every $n\in\mathbb{N}$ there exists $x_n\in A$ such that $$\inf\{|c-x|\mid x\in A\} \leq |c-x_n|\lt \inf\{|c-x|\mid x\in A\}+\frac{1}{n}.$$ What do you know about sequences contained in a compact subset of $\mathbb{R}$?
answered May 13, 2012 at 3:45
$\begingroup$
$\endgroup$
Another approach: If you have learned more general result that any continuous function on a compact space attains its supremum and infimum (see Extreme value theorem at Wikipedia or this question), then it suffices to show that the function $f\colon\mathbb R\to\mathbb R$ defined by $f(x)=|x-c|$ is continuous (for any given $c$).
Showing this is not difficult: You can use triangle inequality or, in the special case of real functions, you can simply notice that it is just shifted absolute value function.
answered May 13, 2012 at 5:03
$\begingroup$
$\endgroup$
0
As Arturo says, all you will have to use, is that a compact space is sequentially compact, that is every convergent sequence in a compact space converges within it!
answered May 13, 2012 at 3:51