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Problem:

Let $X\subset \mathbb{R}^{n}$ be a compact set. Prove that the set $Y=\left \{ y\in \mathbb{R}^{n}: \left | x-y \right |=2000 : x\in X \right \}$ is compact.

First, I don't understand how the absolute value of the difference between x and y: $\left | x-y \right |$ be a number in $\mathbb{R}$, akthough $x$ and $y$ are from $\mathbb{R}^{n}$.

Second, there many definitions for compact set(closed and bounded, every sequence in the set has a convergent subsequence in the set, every open cover has a finite subcover). I am trying to use the first definition, any help please?

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    or if you can show that if $Y$ is closed relative to $\mathbb{R}^n$, then because it is contained in a $k-$ cell that is compact we know that $Y$ will also be compact.2011-12-14

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Here $|x-y|$ means the Euclidean norm $\sqrt{\sum_{j=1}^n (x_j - y_j)^2}$, not absolute value. But your notation $\{y \in {\mathbb R}^n\ :\ |x - y| = 2000\ : \ x \in X\}$ is non-standard. Is the second $:$ supposed to mean "for some"?

In this case, "closed and bounded" is the most useful characterization. But please don't call it a definition, because it doesn't work in other spaces. The fact that subsets of ${\mathbb R}^n$ are compact if and only if they are closed and bounded is a theorem.

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    Yes. So if $y_j$ converges to $y$ and a subsequence of $x_j$ converges to $x$, what do you suppose is $|x - y|$?2011-12-14