Let $E$ and $F$ be two metric spaces. If $K$ is a compact subset of $E$ then a continuous function $f:K\to F$ is always bounded and reachs its maximum.
What happens if we replace $K$ by a closed bounded set $C$ ?
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@900sit-upsaday Sorry, but what's wrong with my question ? – user165633 Aug 17 '14 at 21:45
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bumped this question back up to zero because I honestly see no problem with it. If you're gonna call someone out, it's much better to actually tell them explicitly what was wrong. – syusim Aug 17 '14 at 22:16
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@SamuelYusim Thanks, I am new here, so I would be happy if someone tells me what's wrong. – user165633 Aug 17 '14 at 22:20
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@900sit-upsaday I learned the theorem that says a continuous function on a compact set is bounded, so I asked myself what happens if we consider a bounded closed set instead of compact ? For example this question:http://math.stackexchange.com/questions/6314/is-0-1-a-countable-disjoint-union-of-closed-sets has 44 upvotes despite the fact that it doesn't give any motivation. This is exactly how my question is, it is not that kind of questions where I can motivate the problem. – user165633 Aug 17 '14 at 23:20
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You know what doesn't help someone get answers suited to their needs? Downvoting their question. – syusim Aug 19 '14 at 19:39
1 Answers
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Consider $\mathbb R$ with the metric $d(x,y)=\min\{1,|x-y|\}$. Let $f:\mathbb R \to \mathbb R$ be the function $f(x)=x$. Verify that $\mathbb R$ is closed and bounded and notice that $f$ is not bounded.
Remark: The point is that the "correct" notion of boundedness in metric spaces is that of "total boundedness". For metric spaces a set is compact iff it is complete and totally bounded.
Ittay Weiss
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1Beat me to the punch. This example metric is very useful for finding counterexamples like these. You may want to distinguish between the two copies of $\mathbb{R}$ here. – Dan Rust Aug 17 '14 at 21:24
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