Is the set $\{ x\in \mathbb{Q}: 2< x^2 <3\}$ closed, bounded, compact in $\mathbb{Q}$?

Question

Is the set $\{ x\in \mathbb{Q}: 2< x^2 <3\}$ closed, bounded, compact in $\mathbb{Q}$ ?

I think $\{ x\in \mathbb{Q}: 2< x^2 <3\}=\{ x\in \mathbb{Q}: 2\leq x^2 \leq 3\}$, so it is bounded and closed in $Q$, is that right?

Yes. (Although you might want justification as to why closed intervals are indeed closed in an order topology, which $\Bbb Q$ has.) — anon, Jul 10 '14 at 08:50
Note that your set is not compact because it is not complete (compact iff closed and bounded does not hold for arbitrary spaces). — Matthias Klupsch, Jul 10 '14 at 08:53
@MatthiasKlupsch, I know for the complete space, compact iff closed and bounded, here $\mathbb{Q}$ is not complete. But why the non complete lead to non compact? — Shine, Jul 10 '14 at 08:59
In a compact metric space every sequence has a convergent subsequence. But a Cauchy sequence which is not convergent cannot have a convergent subsequence. — Matthias Klupsch, Jul 10 '14 at 09:06
@mwomath Shouldn't it be both open and closed? Since it contains all its limit points (as $\sqrt{2}, \sqrt{3} \not \in \mathbb{Q}$) it is closed and for every $x$ in the set, there exists a $\varepsilon >0$ such that the ball centered around $x$ with radius $\varepsilon$ is contained in the set, hence open? — Nigel Overmars, Jul 10 '14 at 10:41
Of cousre you are right. I just complete the question, he wrote "in $\mathbb{Q}$" I didn't see it in the first time.
Thanks — Mohammad W. Alomari, Jul 10 '14 at 10:46

mfl · Accepted Answer · 2014-07-10T10:55:52.830

3

The set is not compact. The open cover $U_n=\left((-\sqrt{3},-\sqrt{2})\cup \left(\sqrt{2}+ \frac{1}{n},\sqrt{3}-\frac{1}{n}\right)\right)\cap\mathbb{Q},$ $n\ge 7,$ has no any finite subcover.

edited Jul 10 '14 at 10:55

answered Jul 10 '14 at 10:33

mfl

29,399

score 1 · Answer 2 · answered Jul 10 '14 at 11:35

1

Theorem: Suppose $K\subset Y\subset X$. Then $K$ is compact relative to $X$ if and only if $K$ is compact relative to $Y$.

Using the theorem, if $K$ is compact relative to $\mathbb{Q}$ then it is compact relative to $\mathbb{R}$. Clearly the set is not closed in $\mathbb{R}$.

answered Jul 10 '14 at 11:35

hrkrshnn

6,287

Where did you get such a theorem ? – Learnmore Nov 30 '15 at 16:10
can you provide the source – Learnmore Nov 30 '15 at 16:10
1

@learnmore You can refer to Rudin's Principles of Mathematical Analysis (look into the chapter where he deals with compactness.) – hrkrshnn Dec 01 '15 at 02:40

score 1 · Answer 3 · edited Jul 10 '14 at 11:43

Let $S = \{ x \in \mathbb{Q}_+ : 2 < x^2 < 3\}$.

Let $\{a_n\}_{n=1}^\infty \subset \mathbb{Q}$ be monotonic decreasing and $\lim\limits_{n \to \infty}a_n = \sqrt{2}$ (e.g., $\{1.5; 1.42; 1.415; \ldots\}$)

Let $\{b_n\}_{n=1}^\infty \subset \mathbb{Q}$ be monotonic increasing and $\lim\limits_{n \to \infty}b_n = \sqrt{3}$ (e.g., $\{1.7; 1.73; 1.732; 1.7320; \ldots \}$)

Then, the open cover $\mathcal{O}=(a_n + \dfrac{1}{n}, b_n - \dfrac{1}{n})$ for $n \geq 4$ is an open cover of $S$ from which one cannot extract a finite subcover. Hence it is not compact.

But note that the set is open, closed and bounded in $\mathbb{Q}$.

Is the set $\{ x\in \mathbb{Q}: 2< x^2 <3\}$ closed, bounded, compact in $\mathbb{Q}$?

3 Answers3

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