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I am given the following definition:

Let $B$ be a set of continuous maps with domain a metric space $A$ and codomain a metric space $N$, and $B_x=\{f(x):f\in B\}$.

$B$ is pointwise compact if for each $x\in A, B_x$ is compact in $N$.

I can't figure out what it means for $B_x$ to be compact in $N$. Is the author merely trying to signal that $B_x$ is a subset of $N$? If so, then is this slight rephrasing correct?

$B$ is pointwise compact if for each $x\in A, B_x \subset N$ is compact.

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2 Answers 2

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The intent of the author is probably to make explicit that the topology on $B_x$ is that induced by $N$. Your rephrasing doesn't work: $B$ itself is not a subset of $N$ (just leave out $\subset N$).

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You mean that I can just delete "in N" from the original phrasing, and the rephrasing is correct, right? Thanks. (My original rephrasing was a typo due to haste, and I have changed it.) –  Ryan Sep 29 '13 at 14:54
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Yes, and I would say you can even leave out the other "$\in N$", although that would leave the fact that $B_x$ should get the subspace topology implicit. –  Magdiragdag Sep 29 '13 at 16:17

It is automatic that $B_x$ is a subset of $N$ (by definition of $B_x$); the author means $B_x$ is compact when given the subspace topology (where open sets of $B_x$ are precisely sets of the form $B_x \cap U$, where $U$ is open in $N$).

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If we restrict our study to metric spaces only, then is compactness absolute (not relative to the set it is embedded in)? –  Ryan Sep 29 '13 at 14:53
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Yes, if we understand the metric as given (and we use the metric space topology). I agree that it is better to think of compactness as an absolute notion, if the topology is assumed given. (Being "closed" is not an absolute notion.) So if you understand $B_x$ as automatically carrying the metric inherited from $N$, then you could say this is absolute in that sense. –  user43208 Sep 29 '13 at 15:01
    
Thank you for your clarification. –  Ryan Sep 29 '13 at 15:02

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