# compact sets in $\mathbb{R}^{n}$

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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In this context, $|u|$ stands for the norm (or length) of the vector $u \in \mathbb R^n$. Also, $|x-y|$ is equal to $d(x,y)$, the distance between $x$ and $y$. –  Srivatsan Dec 14 '11 at 4:01
What you’ve written $|x-y|$ is more commonly written $\|x-y\|$, the norm of the vector $x-y$. There are many norms commonly used on $\mathbb{R}^n$; the most familiar is the Euclidean norm, $\|(x_1,\dots,x_n)\|=\sqrt{x_1^2+\cdots+x_n^2}$. That’s the one that I’d assume if another is not explicitly noted. –  Brian M. Scott Dec 14 '11 at 4:04
Note that compactness is not the same as closed and bounded. It is true that compact implies closed and bounded, but the converse is not true in general. However, in $\mathbb R^n$, the converse holds thanks to the Heine-Borel theorem: every closed and bounded subset of $\mathbb R^n$ is compact. –  Srivatsan Dec 14 '11 at 4:07
To prove that the set is closed: since we have finite points of $y$ that satisfies the condition, the our set is a set of points, which implies it is closed. Is that true? How de we prove it is bounded? –  M.Krov Dec 14 '11 at 5:07
As Robert Israel has mentioned, the notation used to describe your set $Y$ is not standard. I don't know what it means. Are you referring to the set of all $y \in \mathbb{R}^n$ such that the distance from $y$ to a fixed point $x$ in $X \subset \mathbb{R}^n$ is 2000? If this is true then we are talking about some $n$ sphere of radius 2000. It is clearly bounded, because it is contained in some $n-$cell. At this point you can look at it one way using the Heine - Borel Theorem, –  user38268 Dec 14 '11 at 7:04

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.
To prove that the set is closed: since we have finite points of $y$ that satisfies the condition, the our set is a set of points, which implies it is closed. Is that true? How de we prove it is bounded? –  M.Krov Dec 14 '11 at 5:07
No, it's not a finite set. One way to show it's closed: suppose a sequence $y_j$ of points of $Y$ converges to $y$. Then there are $x_j \in X$ with $|x_j - y_j| = 2000$. Now what do you know about a sequence in a compact set? –  Robert Israel Dec 14 '11 at 5:21
@Zi2018Alpha Say for example you have a sequence $x_n$ that lives in $X$ which we know is compact. Now the set of all $x_n$ is an infinite set - say we call $E$ - and as a subset of $x_n$ must have a limit point in $X$ because $X$ is compact. –  user38268 Dec 14 '11 at 7:07
Yes. So if $y_j$ converges to $y$ and a subsequence of $x_j$ converges to $x$, what do you suppose is $|x - y|$? –  Robert Israel Dec 14 '11 at 7:56