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I'm trying to prove that $\{0,1,\frac{1}{2},\frac{1}{3},\frac{1}{4},...\}$ is compact in $\mathbb{R}$, but I'm struggling with the definition of compactness. I need to find a finite open subcover, but can I just use the fact that $[0,1]$ is compact? Or is this not even right..

Hope someone can help me with this proof!

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Look at an open cover of $S=\{0\}\cup\{1/n:n\in\Bbb N\}$. It has an open set with $0\in U$. Then $U$ also contains all but finitely many of the $1/n$. So you need only finitely many of the other open sets in the cover to cover the rest of $S$.

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  • $\begingroup$ I'm not sure I fully understand this. My open cover U contains finitely many of the $1/n$, but why the implication that I only need finitely many of the other open sets to cover $S$? $\endgroup$ – jbuser430 Jan 6 '18 at 20:32
  • $\begingroup$ @JBIBB Only finitely many elements of $S$ are not included in the open set $U$ ($U$ contains an open interval around zero). You can afford to be wasteful and cover the remaining elements one-by-one. $\endgroup$ – Jyrki Lahtonen Jan 6 '18 at 20:51
  • $\begingroup$ This is a basic "trick". the point $0$ must be in one open set. But that set is open so it must contain all of the points $\frac 1k$ where $k \ge m$ for some $m$. So that one sets covers ALL but a finite number of $\frac 1k$. The remaining finite number of $\frac 1k$ that are not covered, can each be covered by one different set. So one set can cover $\{\frac 1k, \frac 1{k+1}, ....., 0\}$ and $k-1$ sets can cover $\{ 1, \frac 12, ...., \frac 1{k-1}\}$ so between those $k$ sets, all points are covered. $\endgroup$ – fleablood Jan 6 '18 at 20:52
  • $\begingroup$ Let $\{U\}$ cover $A$. Let $0 \in U_0$. Then $U_0$ is open. Let $0 < \frac 1k < r$ so that $B_r(0) \in U_0$. Then $\{\frac 1k, \frac 1{k+1},\frac 1{k+2},.......,0\} \subset U_0$. For every $m < k$, let $\frac 1m \in U_m$. So $\{1, \frac 12, ....... \frac 1{k-1}\} \subset \cup_{1\lei< k} U_i$ and $\{\frac 1k, \frac 1{k+1},\frac 1{k+2},.......,0\} \subset U_0$. So $A \subset \cup U_0 \cup_{1\lei< k} U_i$. So $\{U_i|0 \le i < k\}$ is a finite subcover. $\endgroup$ – fleablood Jan 6 '18 at 20:58
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Your set is closed and bounded, hence compact.

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    $\begingroup$ That is true, but I'm trying to prove it with the definition that a topological space is compact if from any open cover $U=\{U_i:i\in I\}$ of X one can extract a finite open subcover.. Maybe I wasn't completely clear :) $\endgroup$ – jbuser430 Jan 6 '18 at 20:25
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No, you can't claim that $A =\{0,1,\frac 12, ...\} \subset [0,1]$ is a subset of a compact set is a compact set as there is no reason that subsets of compact sets are compact (because $O$ could be an open cover of $A$ but not be a cover of $[0,1]$).

Let $\{O\}$ be an open cover of $A$. For each $\frac 1k \in A$ there is some $O_{\alpha_k}$ so that $\frac 1k \in O_{\alpha_k}$ and there is some $O_{\tau}$ so that $0 \in O_\tau$.

So $\{O_{\alpha_k}\}\cup \{O_\tau\}$ is a countable subcover of $\{O\}$.

Consid $O_\tau$. It is an open set so that there some $r > 0$ so that $B_r(0) = (-r, r) \subset O_\tau$. Now consider $k > \frac 1r$ then $0 < \frac 1k < r$. So for all $k > \frac 1r$ we have $\frac 1k \in O_\tau$.

So $O_\tau$ is a cover of $\{\frac 1k| k > \frac 1r\}\cup {0}$ And $\{0_{\alpha_k}|k \le \frac 1r\}$ is a finite cover for $\{\frac 1k| k \le \frac 1r\}$.

So $\{O_\tau\} \cup \{0_{\alpha_k}|k \le \frac 1r\}$ is a finite cover for $\{\frac 1k| k > \frac 1r\}\cup {0}\cup\{\frac 1k| k \le \frac 1r\}= A$.

So $\{O\}$ has a finite subcover.

So $A$ is compact.

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  • $\begingroup$ I get it now, thank you so much! $\endgroup$ – jbuser430 Jan 6 '18 at 20:50
  • $\begingroup$ That's a rather misleading answer, because if fails to mention that any closed subset of a compact set is compact, and in the present case, the set in question is indeed a closed subset of $[0,1]$, so the OP can in fact claim as intended. $\endgroup$ – uniquesolution Jan 6 '18 at 21:30
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Let $S = \left\{0,1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\ldots\right\} $.

You can show that it is sequentially compact, i.e. that every sequence in $S$ has a convergent subsequence with it limit in $S$.

A known result from analysis is that if a sequence $(x_n)_{n=1}^\infty$ converges to $x$, then every subsequence $(x_{p(n)})_{n=1}^\infty$ of $(x_n)_{n=1}^\infty$ also converges to $x$. Also, every permutation $(x_{\sigma(n)})_{n=1}^\infty$ where $\sigma : \mathbb{N} \to \mathbb{N}$ is a bijection, also converges to $x$.

Note that the sequence $0, 1, \frac12, \frac13, \frac14\ldots$ converges to $0$.

A sequence in $S$ either has:

  1. Finitely many distinct elements, hence it is eventually constant so it the limit is equal to an element of $S$
  2. Infinitely many distinct elements. Extract one subsequence containing all distinct elements. It is a permuatation of a subsequence of a $0, 1, \frac12, \frac13, \frac14\ldots$ Hence it converges to $0 \in S$ by the above discussion.

Therefore, $S$ is sequentially compact, hence compact.

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We only need that $\frac{1}{n}$ converges to $0$:

If $X$ is a topological space, and $x_n \to x$ is a convergent sequence in $X$, then $K=\{x_n: n \in \mathbb{N}\} \cup \{x\}$ is compact.

Let $\mathcal{U}$ be an open cover of $K$. There must be some $U_x \in \mathcal{U}$ such that $x \in U_x$. Then $U_x$ is an open neighbourhood of $x$ so the definition of convergence tells us that there is some $N \in \mathbb{N}$ such that $$\forall n \ge N: x_n \in U_x$$

Now for each $i \in \{1,\ldots, N-1\}$ pick some $U_i \in \mathcal{U}$ such that $x_i \in U_i$.

Then $\{ U_1, \ldots, U_{N-1}, U_x\}$ is a finite subcover of $\mathcal{U}$: all elements $x_n$ are covered by $U_n$ if $n < N$ and by $U_x$ if $i \ge N$, and $x \in U_x$ as well.

A convergent sequence with its limit is, in a way, the prototypical compact set.

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