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# sequentially compact watch

1. You have the metric space N (natural numbers) with distance measured with d(x,y)=|(1/m) -(1/n)|. I have proved that it's a metric space, and i have also shown that every subset of N is open and closed. I need to show that any infinite subset X of N is not sequentially compact. So i know i nedd to show that there exists sequence in X that converges to a point not in X but i don't know how to?
2. (Original post by Ishika)
You have the metric space N (natural numbers) with distance measured with d(x,y)=|(1/m) -(1/n)|. I have proved that it's a metric space, and i have also shown that every subset of N is open and closed. I need to show that any infinite subset X of N is not sequentially compact. So i know i nedd to show that there exists sequence in X that converges to a point not in X but i don't know how to?
Do you have the equivalence of compactness and sequential compactness for a metric space? If so, you can take singletons as the open cover and as the subset is infinite, the cover doesn't have a finite subcover.

Otherwise, you need to find a sequence in X with no convergent subsequence. The only sequences that converge are those that are eventually constant. Find a sequence in X with the property that no subsequence becomes constant.
3. How do you know that the sequences which converge are eventually constant, because i can't get to that with the definition of convergence; i just get |(1/m) -(1/n)|< e (epsylon)
4. (Original post by Ishika)
How do you know that the sequences which converge are eventually constant, because i can't get to that with the definition of convergence; i just get |(1/m) -(1/n)|< e (epsylon)
If L is the limit of a convergent sequence (x_1, x_2, ...), you can find e>0 so that |(1/L)-(1/k)|<e for all k not equal to L. But there is some N so that n>N => |(1/L)-(1/(x_n))|<e. So the only option is that x_n = L for all n > N.

Or, you can use an equivalent definition of convergence in a metric space (in fact the definition for a topological space). If L is the limit then for any open set U containing L, there is an N so that n>N => x_n is in U. So take U to be {L}.
5. I don't know if it helps your thinking, but you could also note that your space is isometric to

{1, 1/2, 1/3, 1/4, ...}

with the usual metric on the reals.
6. (Original post by SsEe)
If L is the limit of a convergent sequence (x_1, x_2, ...), you can find e>0 so that |(1/L)-(1/k)|<e for all k not equal to L. But there is some N so that n>N => |(1/L)-(1/(x_n))|<e. So the only option is that x_n = L for all n > N.
I don't really understand the last part. How do you get
that x_n =L eventually?
7. (Original post by Ishika)
I don't really understand the last part. How do you get
that x_n =L eventually?
It's from how I defined e. e is such that |(1/L)-(1/k)|<e => k=L. Take e to be epsilon in the definition of a convergent sequence.

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