MHB Problem of the Week #111 - July 14th, 2014

[Chris L T521]

Here's this week's problem!

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Problem: Show that a connected metric space having more than one point is uncountable.

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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Chris L T521]

This week's problem was correctly answered by Euge and magneto. You can find Euge's solution below.

[sp]Let $(M, d)$ be a connected metric space with more than one point. Take distinct points $x, y\in M$, and define a function $f : M \to [0,1]$ by setting

$f(a) = \dfrac{d(x, a)}{d(x, a) + d(a, y)}.$

By the triangle inequality, $d(x, a) + d(a, y) \ge d(x, y) > 0$ for all $a \in M$. So $f$ is well-defined. Since the maps $a \mapsto d(x, a)$ and $a \mapsto d(a, y)$ are continuous, $f$ is continuous. Furthermore, $f(x) = 0$ and $f(y) = 1$. For $0 < k < 1$, the intermediate value theorem gives a $c\in M$ such that $f(c) = k$. Consequently, $f$ maps $M$ onto $[0,1]$, an uncountable set. Therefore $M$ is uncountable.[/sp]