1. ## compactness

Regard Q, the set of all rational numbers, as a metric space, with
d(p,q)=|p-q|. Let E be the set of all P in Q such that 2<P^2<3. Show
that E is closed and bounded in Q, but that E is not compact.
Is E open in Q?

I could prove that E is closed, open and bounded. But will the following solution suffice to show that E is not compact?

I can find a sequence of points {Kn} = 1.7, 1.73, 1.732,.... in E. This is a sequence converging to 3^(1/2).
So {Kn} is an infinite subset of E which does not have a limit point in E.
So E is not compact.

2. I can find a sequence of points {Kn} = 1.7, 1.73, 1.732,.... in E. This is a sequence converging to 3^(1/2).
So {Kn} is an infinite subset of E which does not have a limit point in E.
So E is not compact.
I think that you need to be careful about your claim that $\{K_n\}$ does not have a limit point in $E$. How do you know?

3. @ roninpro
Let x be a limit point of {Kn}. Then I can find a sequence of points in Kn converging to x. But {Kn} = 1.7, 1.73, 1.732, ..... which is a seq convering to 3^(1/2), so any subsequence also converges to 3^(1/2). So x has to be 3^(1/2).

Am I right?

4. Yes, but the point is a little more subtle than that. Because your sequence, as a sequence of real numbers, converges to $\sqrt{2}$ and limits are unique, it cannot converge to any other real number and so, since the rationals are a subset of the reals, it cannot converge to any rational number. Now, that is sufficient to say that your sequence, as a seuquence in the rational numbers, does not converge.

5. Originally Posted by poorna
Regard Q, the set of all rational numbers, as a metric space, with
d(p,q)=|p-q|. Let E be the set of all P in Q such that 2<P^2<3. Show
that E is closed and bounded in Q, but that E is not compact.
Is E open in Q?

I could prove that E is closed, open and bounded. But will the following solution suffice to show that E is not compact?

I can find a sequence of points {Kn} = 1.7, 1.73, 1.732,.... in E. This is a sequence converging to 3^(1/2).
So {Kn} is an infinite subset of E which does not have a limit point in E.
So E is not compact.
As you have it we have that $\left(\mathbb{Q},d\right)$ is a subspace of $\mathbb{R}$ with the usual topology. Now, it is true that if $X$ is a topological (specifically metric space) and $Y$ a subspace then $Z$ is a compact subspace of $Y$ iff it is a compact subspace of $X$. So, now as you noted $E$ would have to be compact in $\mathbb{R}$ to be compact in $\mathbb{Q}$ but it is definitely not for the simple reason that $x_n=\frac{1}{2}\left(x_n+\frac{2}{x_n}\right),\tex t{ }x_0$ is a sequence of points in $E$ such that $x_n\to \sqrt{2}$ and so $\sqrt{2}$ is a limit point of $E$ but not a point of $E$...in other words $E$ isn't even a closed subspace of $\mathbb{R}$! What does the Heine-Borel theorem say?

6. @ Drexel28
Yeah, E has to be a closed and bounded subset of R in order to be compact in R. And ince compactness i independent of the set in which is contained, we have that E is not compact.
Thanks!