# Thread: Prove that A is compact

1. ## Prove that A is compact

A={0}u{1/n: n is a positive integer}. Prove A is a compact subset of Q w.r.t the standard topology but non-compact w.r.t discrete topology

2. ## Re: Prove that A is compact

Well, what have you tried? At least tell us what definition of "compact" you are using. There are several definitions that are equivalent under specified conditions.

3. ## Re: Prove that A is compact

X is compact if every open cover of X has a finite subcover. I don't know what to do.

4. ## Re: Prove that A is compact

A hint for the first one is that any open set that contains 0 contains infinitely many points of A. For the second one, note that the set of the open sets {1/n} and {0} have no finite subset which contains all the elements of A.

5. ## Re: Prove that A is compact

Suppose we have an open cover for this set. Then there exist a set, U, in the cover containing 0. That set, because it is open, must contain some open interval $\displaystyle (-\delta, \delta)$. The sequence 1/n converges to 0. In particular, that means that there exist N such that [te]0< 1/N< \delta[/tex] so all members of the sequence beyond 1/N are in U. Now just choose one set in the cover for each $\displaystyle n\le N$. That gives a finite number of sets from the open cover that cover the set.

If you have the theorem that "A subset of the real numbers is compact if and only if it is both closed and bounded", then it is easy to see that the set is bounded because it is a subset of [0, 1] and that it is closed because the only limit point is 0 which is in the set.

6. ## Re: Prove that A is compact

to see it is NOT compact w.r.t. the discrete topology, note that {{x}: x is in A} is an open cover of A (since {x} is open in the discrete topology, since EVERY subset of A is open in the discrete topology) which has NO finite subcover, since A is not a finite set.

7. ## Re: Prove that A is compact

In the discrete topology a set is compact if and only if it is finite. This set is not finite.
(The proof that if a set if compact, with the discrete topology, then it is finite is esentially what Deveno gave.

8. ## Re: Prove that A is compact

Halls, I thought Heine-Borel only applied to R?

9. ## Re: Prove that A is compact

a version of Heine-Borel can be constructed in any metric space:

a set K in a metric space is compact (using the open cover definition) if and only if it is closed, and totally bounded (we have to use totally bounded, rather than just bounded, because the discrete metric on an infinite set otherwise provides a counter-example).

in particular, in the vector space Rn, with the metric induced by the euclidean inner product ("dot product"), a subset (with the relative topology) is totally bounded, if and only if it is bounded.

the necessity of closure in the definition is seen by considering the set S = {x in Rn: ||x|| < 1} along with the open cover O = {B(0,(n/(n+1)): n in N} where:

B(0,y) = {x in Rn: ||x|| < y}. the necessity of bounded is seen by considering the set Rn itself, and the open cover O = {B(0,n): n in N}.

i remark in passing that the standard topology on R is the one induced by the metric d(x,y) = |x - y|, which is the same one we get by considering R as a 1-dimensional euclidean vector space:

in 1 dimension, the dot product is simply real multiplication, and the norm induced by this is: ||x|| = √(x.x) = √(x2) = |x|, the absolute value of x.

topology is all about "nearness" (or the lack of it), having a distance function makes this easy to define (without it, we have to settle for "neighborhood", because we have no other yardstick to measure by).

the indiscrete topology is like viewing a set from so far away, it seems like a single point. the discrete topology is like being so close to every point, we can't even imagine another one near-by (no more "near" and "far", only "here" and "not-here").

your original problem indicates the usual (standard = metric) topology is being used on R, so Heine-Borel would indeed apply.