Thread: 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.

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 does not have a limit point in . How do you know?

@ 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?

Yes, but the point is a little more subtle than that. Because your sequence, as a sequence of real numbers, converges to 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.

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 is a subspace of with the usual topology. Now, it is true that if is a topological (specifically metric space) and a subspace then is a compact subspace of iff it is a compact subspace of . So, now as you noted would have to be compact in to be compact in but it is definitely not for the simple reason that is a sequence of points in such that and so is a limit point of but not a point of ...in other words isn't even a closed subspace of ! What does the Heine-Borel theorem say?

@ 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!