Show that $S=\mathbb{Q} \cap [0,2]$ is not compact

Question

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.

Answer 1

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

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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 $$ \lim_{n\to\infty}|a_n^2-2|=0\tag{5} $$

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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, $$ \mathbb{Q}\cap[0,2]\text{ is not compact.}\tag{8} $$

Answer 2

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).

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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 $\ddot\smile$

Answer 3

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.