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In $\mathbb{R}^n$ we prove that a set is compact (using the definition about open covers) if and only if it's closed and bounded. It is pretty clear that if $\mathcal{O}$ is an open cover of one unbounded set $X$, then $\mathcal{O}$ cannot have a finite subcover, it'll clearly need in general infinitely many sets to cover the set $X$.

Now, if a set isn't closed, I cannot see in which way it fails to be compact. For instance, if $X$ is the closed unit ball centered at the origin, then it is compact. If on the other hand we consider $Y=X\setminus\{0\}$, then it's not compact anymore, because $Y$ isn't closed (the point $0$ is a limit point of $X$ and so, $0 \in \operatorname{Cl}(X)$ and on the same time $0 \notin Y$.

So, what should be the intuition about this? How can we intuitively see that $Y$ isn't compact?

Thanks very much in advance!

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4 Answers 4

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Let $p$ be a limit point of the set $Y$ that is not contained in $Y$. Then $(U_n)_{n\in\mathbb{Z}^+}$ with

$$U_n := \{ y \in Y : \lvert y - p\rvert > \frac1n\}$$

forms an open cover of $Y$ which has no finite subcover.

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Two equivalences which are true for metric spaces are these:

  1. $A$ is compact if and only if every sequence of elements from $A$ has a convergent subsequence with limit in $A$.
  2. $A$ is closed if and only if every convergent sequence of elements from $A$ has its limit in $A$.

Now it is clear why a set which isn't closed is not compact. If $A$ is not closed, then there is some sequence $a_n$ of elements of $A$, whose limit is $a\notin A$. But the uniqueness of limits mean that every subsequence of $a_n$ converges to $a$. Therefore there is no convergent subsequence whose limit lies in $A$.

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  • $\begingroup$ That's the definition of (sequential) compactness. I don't understand your question... $\endgroup$
    – Asaf Karagila
    Feb 29, 2016 at 15:45
  • $\begingroup$ I misunderstood...you cleared up my confusion without understanding my question ...thanks :) $\endgroup$
    – JKnecht
    Feb 29, 2016 at 16:06
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Since this question is tagged (general-topology), it may be good to note that in a general topological space compact sets need not be closed. A simple counterexample is obtained by taking any finite space with a non-discrete topology. All subsets of such a space are trivially compact (any open cover without repeated sets must be finite), but by assumption not all subsets are closed.

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    $\begingroup$ Here is one very stupid example I like of compact sets being not necessarily closed: Take $\Bbb{R}$ with the trivial topology and $(0,1)$ is compact. $\endgroup$
    – user38268
    Aug 24, 2013 at 16:20
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    $\begingroup$ I myself prefer the co-finite topology on an infinite set. Because it's a topology which rises naturally in mathematics, and it's pretty cool too. $\endgroup$
    – Asaf Karagila
    Aug 24, 2013 at 17:28
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Another way to see this is to recall that compactness is defined via open sets, and thus is a topological property. So for a space $X$ and another space $Y$ that is homeomorphic to $X$ then $X$ is compact if and only if $Y$ is as well.

Now consider your "punctured ball" example. We will assume that we are dealing with $\mathbb{R}^2=\mathbb{C}$ for simplicity. Then the puncture ball is homeomorphic to the complement of the open unit ball (ie to the set $\{z=x+iy: |z|\geq1\}$) via the map $z\mapsto\frac{1}{z}$. This last set is unbounded and so by your intuition should not be compact. In fact you can more or less always do this for a non-closed set (meaning you take a point in the closure and "send it to infinity" and get an unbounded set that is homeomorphic to your original set.

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