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I'm self-studying from the book Understanding Analysis by Stephen Abbott, and I have a question about the prove of theorem 3.3.4 on page 84 (i.e. the Heine-Borel theorem).

To be more specific, let us assume $K \subseteq \mathbb{R}$ to be compact.$^1$ Then we would like to prove that this implies that $K$ is closed.$^2$ Stephen Abbott proceeds by considering a sequence $(x_n)$ with $x = \lim x_n$, where $(x_n)$ is contained in $K$. By definition of a compact set, the sequence $(x_n)$ has a convergent subsequence $(x_{n_k})$, and it follows from a well-known theorem$^3$ that $(x_{n_k})$ converges to the same limit $x$. Finally, the definition of a compact set requires that $x \in K$. Then according to Stephen Abbott, this proves that $K$ is closed.

But in his prove he has assumed that the sequence $(x_n)$ converges. What happens if $(x_n)$ does not converge? Then we can still have a subsequence that converges to a limit, right? If so, then I think his theorem is incomplete, or am I missing something?

$^1$ A set $K \subseteq \mathbb{R}$ is compact if every sequence in $K$ has a subsequence that converges to a limit that is also in $K$.

$^2$ A set $K \subseteq \mathbb{R}$ is closed if it contains its limit points.$^{2.1}$

$^{2.1}$ A point $x$ is a limit point of a set $K$ if every $\epsilon$-neighborhood $V_\epsilon (x)$ of $x$ intersects the set $K$ in some point other than $x$.

$^3$ Subsequences of a convergent sequence converge to the same limit as the original sequence.

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  • $\begingroup$ Just a note about terminology, technically your $^1$ is called sequentially compact. In general topological spaces this doesn't necessarily imply compact nor compact imply this. $\endgroup$
    – DanZimm
    Aug 25, 2014 at 11:54
  • $\begingroup$ If u ve a sequence that does not converge then u dont have to show anything. $\endgroup$
    – FWE
    Aug 25, 2014 at 12:53
  • $\begingroup$ @YonedaLemma could you please elaborate on that? Because that is exactly what I am confused about. $\endgroup$
    – Hunter
    Aug 25, 2014 at 12:56
  • $\begingroup$ @YonedaLemma For the proof, his first assumption is that he is considers a sequence $(x_n)$ with $x= \lim x_n$. But what if we consider a sequence $(x'_n)$ that does not converge. As far as I know, we can still have a convergent subsequence $(x'_{n_k})$, but now we cannot use the theorem that "subsequences of a convergent sequence converge to the same limit as the original sequence", and so his prove breaks down. I must be missing something, but I don't see what it is. $\endgroup$
    – Hunter
    Aug 25, 2014 at 12:56
  • $\begingroup$ $K$ is closed iff every convergent seq. in $K$ converges against a point in $K$. So u have to show: If $(x_n)$ is in $K$ and conv. against some $x$ then $x$ is in $K$. How: If $(x_n)$ is a seq. in $K$ that converges then because $K$ is compact, $(x_n)$ has a convergent subseq. with limit in $K$. By the theorem u mentioned the limit of the subsequent is identical to the limit of the original $(x_n)$ - so it must be in $K$ $\diamond$. (If for a proof u start with a seq. that does not conv. then u are just not about to prove that $K$ suffices the given definition of being a closed set). $\endgroup$
    – FWE
    Aug 25, 2014 at 13:03

2 Answers 2

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I think you are not understanding what is going on. You need to prove that $K$ is closed, so you must prove that any point $x$ that is the limit of a sequence of pints $x_n \in K$ belongs to $K$. So you must assume that $x=\lim_n x_n$!

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  • $\begingroup$ Yeah, I am still confused because that is not the definition of a closed set, as far as I understand it. Is there a theorem that every sequence in a closed set must be convergent? $\endgroup$
    – Hunter
    Aug 25, 2014 at 11:07
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    $\begingroup$ Such a theorem would be false... The author is using this definition of closedness: a set $C$ (in a metric space) is closed if $x \in C$ whenever $x=\lim_n x_n$ with $x_n \in C$ for every $n$. In other words, $C$ contains its cluster points. $\endgroup$
    – Siminore
    Aug 25, 2014 at 11:12
  • $\begingroup$ I'm really sorry, but I'm still confused. For the proof, his first assumption is that he is considers a sequence $(x_n)$ with $x= \lim x_n$. But what if we consider a sequence $(x'_n)$ that does not converge. As far as I know, we can still have a convergent subsequence $(x'_{n_k})$, but now we cannot use the theorem that "subsequences of a convergent sequence converge to the same limit as the original sequence", and so his prove breaks down. I must be missing something, but I don't see what it is. $\endgroup$
    – Hunter
    Aug 25, 2014 at 12:55
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    $\begingroup$ I think I understand it, but I'm not sure. In order to prove that $K$ is closed, we must show that $K$ contains its limit points. A well-known theorem$^1$ tells us that every limit point $x$ of $K$ corresponds to some convergent sequence $(x_n)$ contained in $K$, that is, $x = \lim x_n$. Then he subsequently uses that to prove that $K$ is closed if $K$ is compact, as I have described above. --- $^1$ Theorem: A point $x$ is a limit point of a set $K$ if, and only if, $x = \lim x_n$ for some sequence $(x_n)$ contained in $K$ satisfying $x_n \neq x$ for all $n \in \mathbb{N}$. $\endgroup$
    – Hunter
    Aug 25, 2014 at 13:37
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    $\begingroup$ Yes, your last interpretation is fine. $\endgroup$
    – Siminore
    Aug 25, 2014 at 16:06
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Definition (Hausdorff space)

A topological space $X$ is called Hausdorff or $T_2$ space if and only iff every two points $x,y\in X$ have disjunct neighbourhoods $x\in O_x\subset x$ and $y\in O_y\subset X$.

Example (Metric spaces are Hausdorff)

Every metric space $M$ is (as a topological space) $T_2$ - for each $x, y\in M$ say $d:=d(x,y)$ is their distance, then the $\epsilon$-spheres around $x$ and $y$ with $\epsilon < \frac{d}{2}$ (f.i. open intervals with midpoint $x$ resp. $y$) are disjoint neighbourhoods of $x$ and $y$. Therefore especially $R$ is $T_2$ (since $R$ is a metric space).

Proposition (Compact subsets of $T_2$ spaces are closed)

Say $A\subset X$ where $X$ is a topological space that is Hausdorff (f.i. $R$).

If $A$ is compact then $A$ is closed.

Proof: Say $X$ is $T_2$ and $A\subset X$ is compact. We show that $A^c$ is open (and thereby $A$ is closed) by showing that any $y\in A^c$ has a neighbourhood $V_y\subset A^c$:

Say $y\in A^c$ arbitrary. Then for any $x\in A$ we have a neighbourhood $O_x$ of $x$ and a neighbourhood $V_y^x$ of $y$ that is disjoint to $O_x$ (since $X$ is $T_2$). So we have $$A\subset \bigcup_{x\in A}O_x$$ and $$\bigcup_{x\in A}O_x \cap\bigcap_{x\in A}V_y^x=\emptyset$$

and therefore $$\bigcap_{x\in A}V_y^x\subset A^c$$

Because $A$ is compact we can argue with finitely many $x\in A$. But then $V_y:=\bigcap_{x\in A}V_y^x$ is open - so altogether $V_y$ is an open neighbourhood of $y$ that is contained in $A^c$. $\diamond$

(For $X=R$ the $O_x$ and $V_y^x$ become open intervals in $R$ with midpoint $x$ resp. $y$. )

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  • $\begingroup$ Thanks for your answer. Unfortunately this is too advanced for the book I'm currently reading. The author has not defined what a topological space or a metric space is. (But don't delete the answer, because it might be useful for someone else.) $\endgroup$
    – Hunter
    Aug 25, 2014 at 13:02
  • $\begingroup$ Yes I see - you are just confused about some simple point namely what to show for the proof. When besides the things the author shows u want also to show something else (f.i. for non converging sequences in $K$) then u are trying to do too much - see above comment. $\endgroup$
    – FWE
    Aug 25, 2014 at 13:11
  • $\begingroup$ I think I understand it with your help. In order to prove that $K$ is closed, we must show that $K$ contains its limit points. A well-known theorem$^1$ tells us that every limit point $x$ of $K$ corresponds to some convergent sequence $(x_n)$ contained in $K$, that is, $x = \lim x_n$. Then he subsequently uses that to prove that $K$ is closed if $K$ is compact, as I have described above. --- $^1$ Theorem: A point $x$ is a limit point of a set $K$ if, and only if, $x = \lim x_n$ for some sequence $(x_n)$ contained in $K$ satisfying $x_n \neq x$ for all $n \in \mathbb{N}$. $\endgroup$
    – Hunter
    Aug 25, 2014 at 13:36
  • $\begingroup$ Can you double check if the above is correct? $\endgroup$
    – Hunter
    Aug 25, 2014 at 13:36
  • $\begingroup$ Not quite - a limit point of a set is a point that some sequence in that set converges to (usually that is taken as definition - not as well known theorem). A limit point of a set therefore corresponds not to a sequence in that set that converges to it but in general to any sequence in the set that converges to it. $\endgroup$
    – FWE
    Aug 25, 2014 at 13:42

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