2
$\begingroup$

Question: Let $S$ be the set of rational numbers in the interval $[0, 2]$. Using the definition of compactness, show that $S$ is not compact.

Using the definition of closeness/sequential-compactness, it is easy to show that $S$ is not closed/sequential-compact, and thus it is not compact.

But I am trying to show non-compactness of $S$ by a counterexample using purely definition of compactness, but it is not possible; since every covering example for $A=[0, 2]$ seems to be the same for $S=\mathbb{Q} \cap A$.

I highly appreciate some guidance.

EDIT Answers in here are either relevant to Topology or using non-closeness as an intermediate.

$\endgroup$
  • 7
    $\begingroup$ Try an open cover consisting of sets of the form $(S\cap[0,\sqrt2-1/n))\cup(S\cap(\sqrt2+1/n,2])$. $\endgroup$ – David Mitra Jan 31 '15 at 10:05
  • 1
    $\begingroup$ @David Mitra: Thank you very much. Just a bit question: Is it $(-0.1,\sqrt2-1/n)\cup(\sqrt2+1/{(2n)},2.1)$? $[]\rightarrow ()$ because it must be (?) open intervals, and $n\rightarrow 2n$ because of definition of interval? $\endgroup$ – L.G. Jan 31 '15 at 10:17
  • $\begingroup$ You can work with those sets also. Probably better than what I suggested (where I thought of the ambient space as $S$ with the subspace topology). $\endgroup$ – David Mitra Jan 31 '15 at 10:26
  • 3
    $\begingroup$ "every covering example for $A=[0, 2]$ seems to be the same for $S=\mathbb{Q} \cap A$." This is true but one should start from some covering of $S=\mathbb{Q} \cap A$. Since some them are not covering $A$, this makes all the difference. $\endgroup$ – Did Jan 31 '15 at 10:48
  • $\begingroup$ You might want to be a bit clearer about which definition of compactness you want us to use. It takes a bit of effort to determine that. $\endgroup$ – robjohn Jan 31 '15 at 15:32
3
$\begingroup$

Although there is nothing wrong with proofs that do, I tried not to reference irrational numbers in the following proof.


Define a Sequence of Squares Converging to $\boldsymbol{2}$

Consider the sequence $$ a_1=1\quad\text{and}\quad a_{n+1}=\frac{a_n}2+\frac1{a_n}\tag{1} $$ If $a_n\in[1,2]$, then $(1)$ implies $a_{n+1}\in[1,2]$.

Each $a_n$ is in $\mathbb{Q}\cap[0,2]$ and $$ \begin{align} a_{n+1}^2-2 &=\frac{a_n^2}4-1+\frac1{a_n^2}\\ &=\left(\frac{a_n^2-2}{2a_n}\right)^2\tag{2} \end{align} $$ For $x\gt0$, $\dfrac{x^2-2}{2x}=\dfrac x2-\dfrac1x$ is monotonically increasing. Therefore, for $a_n\in[1,2]$ we have $$ -\frac12\le\frac{a_n^2-2}{2a_n}\le\frac12\tag{3} $$ Combining $(2)$ and $(3)$ gives $$ \begin{align} \left|a_{n+1}^2-2\right| &=\left(\frac{a_n^2-2}{2a_n}\right)^2\\ &=\left|\frac{a_n^2-2}{2a_n}\right|\frac1{2a_n}\left|a_n^2-2\right|\\ &\le\frac14\left|a_n^2-2\right|\tag{4} \end{align} $$ which implies $$ \lim_{n\to\infty}|a_n^2-2|=0\tag{5} $$


Define an Infinite Cover of $\boldsymbol{\mathbb{Q}\cap[0,2]}$

Consider the open sets $$ U_n=\left\{x\in\mathbb{Q}\cap[0,2]:|x^2-2|\gt\frac1n\right\}\tag{6} $$

According to this answer there is no rational number so that $x^2=2$; therefore, we have $$ \bigcup_{n=1}^\infty U_n=\mathbb{Q}\cap[0,2]\tag{7} $$ Since $\bigcup\limits_{n=1}^mU_n=U_m$, the union of any finite collection of $U_n$ is another $U_n$. $(5)$ guarantees that no matter how big we choose $n$, there is an $a_{k_n}\not\in U_n$. Since $a_{k_n}\in\mathbb{Q}\cap[0,2]$, there is no $U_n$ that contains all of $\mathbb{Q}\cap[0,2]$.

That is, no finite subset of $\{U_n\}$ can cover all of $\mathbb{Q}\cap[0,2]$. Therefore, $$ \mathbb{Q}\cap[0,2]\text{ is not compact.}\tag{8} $$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I have updated the answer a bit to use the desired definition of compactness. $\endgroup$ – robjohn Jan 31 '15 at 18:10
2
$\begingroup$

Disclaimer: Somehow I've missed the comment of David Mitra; as my answer is essentially the same, I'm making this CW. I'm not deleting because here $U_i\cap U_j = \varnothing$, so peraps it is more clear that no finite cover exists (all the sets $U_n$ have to be taken).


Take any strictly decreasing sequence $(a_n)_{n=1,2,\ldots}$ of non-rational numbers that tends to $\sqrt{2}$ with $a_1 < 3$, for example

$$a_n = \sqrt{2}+\frac{1}{n+2015}$$

Now consider a sequence of open sets:

\begin{align} U_0 &= (-1,\sqrt{2})\\ U_1 &= (a_1,3)\\ U_n &= (a_n,a_{n-1}) &\text{ for } n \geq 2 \end{align}

Obviously $\mathbb{Q}\cap [0,2] \subset \bigcup_{k=1}^{\infty} U_k$, but no finite cover exists.

I hope this helps $\ddot\smile$

$\endgroup$
0
$\begingroup$

A compact set has to be bounded and closed. We can easily see that closedness is violated here. For example, take the famous Leibniz formula $$\frac{\pi}{4}= 1-\frac13+\frac15-\frac17+\frac19-\cdots$$ The sequence of partial sums are all rational numbers in between 0 and 2, however their limit point is not rational, so the set $S$ is not closed.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.