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We have $\Bbb Q$ equipped with the Euclidean Metric.

Show that $A=([0,\sqrt2] \cap \Bbb Q ) \subset \Bbb Q $ is not compact.

How would you go about showing this?

You can make on open cover $\{O_n\}= ((-\infty, \sqrt2 -\frac1n) \cap \Bbb Q)$ for $n\ge1$

$ A \subset {O_n}$

How would you show there is no finite subcover?

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Hint:

What can you say about the sequence (and its subsequences)

$$\left\{\left(1-\frac1n\right)^n\right\}_{n\in\Bbb N}\subset A\;\;?$$

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Suppose there was a finite subcover. Then $A \subset (-\infty, \sqrt2 -\frac1n) \cap \mathbb{Q}$ for some $n$. However, $[\sqrt2 -\frac1n, \sqrt{2} ] \cap \mathbb{Q} \neq \emptyset$, which contradictions the definition of $A$.

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  • $\begingroup$ Are you saying that $n$ is finite? $\endgroup$ – user113867 Dec 20 '13 at 17:56
  • $\begingroup$ Yes, if $O_{n_1},...,O_{n_k}$ is finite open cover of $A$, then let $n = \max(n_1,...,n_k)$. $\endgroup$ – copper.hat Dec 20 '13 at 17:58
  • $\begingroup$ @copper.hat I upvoted. Your answer answers the OP's question. $\endgroup$ – Stefan Smith Dec 20 '13 at 21:01
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Not sure, whether you already know that images of compact spaces under continuous maps are compact, but in case you do, this criterion may be sometimes easier to use than coming up with open covers. In particular, any continuous function on a compact attains its extremal values. Let $X := \mathbb Q\cap[0,\sqrt2]$ and define $\iota:X\to \mathbb R$ be the embedding map. Clearly, it is continuous and clearly $\sup_{x\in X}\iota(x) = \sqrt 2$, however for no $x\in X$ it holds that $\iota(x) = \sqrt2$, so $X$ is not compact.

Perhaps, in the current situation it is not much different from other solutions provided, however I think this trick is just worth keeping in mind for the future problems in topology.

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If a subset $A$ of $\mathbb R$ is compact, then it is also closed. In particular, if a ssequence $\{x_n\}_{n\in}\subset A$ converges to $x$, then $x\in A$. Let \begin{align} x_1=&1\\ x_2=&1.4 \\ x_3=&1.41 \\ x_4=&1.414 \\ x_5=&1.4142 \\ etc. \end{align} I.e., $$ x_n=\frac{\lfloor10^n\sqrt{2}\rfloor}{10^n}, $$ Clearly $x_n\to\sqrt{2}$, as $|x_n-\sqrt{2}|<10^{-n}$, and $\sqrt{2}\not\in A$. Hence $A$ is not compact.

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  • $\begingroup$ Essentially, you seem to use the same approach as I did without explicitly mentioning it. I believe you shall prove the relevance of If a subset of $\Bbb R$ is compact, then it is also closed to the setting of OP. $\endgroup$ – Ilya Dec 20 '13 at 18:22
  • $\begingroup$ What is that notation you are using for the numerator? When I tried to create a sequence like this I could only come up with $(\sqrt2 - \frac1n $ but this has elements $\notin A$ . I was looking for a way to come up with something like this! $\endgroup$ – user113867 Dec 20 '13 at 20:44
  • $\begingroup$ It is the integer part of $10^n\sqrt{2}$. $\endgroup$ – Yiorgos S. Smyrlis Dec 20 '13 at 20:46
  • $\begingroup$ @user113867: it is handy to remember that any real number can be approximated by rational ones (that's actually can be considered as one of definition of reals): just think of the decimal expression of real numbers. Whenever you truncate it, you obtain a rational number which better and better approximates the real one. This truncation is what Yiorgos wrote explicitly here. $\endgroup$ – Ilya Dec 20 '13 at 22:46
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It is not sequential compact hence not compact

$x_n={[\sqrt{2}n]\over n}\in [0,\sqrt{2}]\cap\mathbb{Q},x_n\to\sqrt{2}\notin [0,\sqrt{2}]\cap\mathbb{Q}$

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  • $\begingroup$ But $x_1 = \sqrt2 \notin [0, \sqrt2] \cap \Bbb Q $ $\endgroup$ – user113867 Dec 20 '13 at 17:46
  • $\begingroup$ $x_1=1$.................... $[x]=x-\{x\}$ where $\{x\}$ is fractional part of $x$ $\endgroup$ – Marso Dec 20 '13 at 17:55
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Let $O_{n_1},\ldots,O_{n_k}$ be a finite subcover, with $n_1<\cdots<n_k$. Then $$\left(\sqrt2-\frac{1}{n_k},\sqrt2-\frac{1}{n_k+1}\right)\cap \mathbb Q$$ is nonempty and contained in $[0,\sqrt2]\cap \mathbb Q$ but disjoint from $O_{n_1}\cup\cdots\cup O_{n_k}$, so $[0,\sqrt2]\cap \mathbb Q\nsubseteq O_{n_1}\cup\cdots\cup O_{n_k}$.

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