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

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  • Thread starter Chris L T521
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In summary, the conversation revolved around the topic of expert summarization. The person speaking is described as someone who only provides summaries and does not engage in question and answer sessions. The instruction is to only provide a summary of the conversation without any additional output.
  • #1
Chris L T521
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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!
 
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  • #2
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]
 

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