- #1
Cairo
- 61
- 0
Let X be an infinite set and p be a point in X, chosen once and for all. Let T be the collection of open subsets V of X for which either p is not a member of V, or p is a member of V and its complement ~V is finite.
Now, let (a_n) be a sequence in X (that is, for all n in N, a_n in X) such that the set of the sequences, {a_n : n in N}, is infinite. Using the definition of convergence in topological spaces, prove that (a_n) has a subsequence which converges to p.
I thought I could do this by showing that if a sequence (a_n) converges to p, then so does every subsequence. But then realized that (a_n) is ANY sequence, so my proof would not hold. I'm also not sure if a sequence even does converge to p!
Any ideas how to prove this result?
Now, let (a_n) be a sequence in X (that is, for all n in N, a_n in X) such that the set of the sequences, {a_n : n in N}, is infinite. Using the definition of convergence in topological spaces, prove that (a_n) has a subsequence which converges to p.
I thought I could do this by showing that if a sequence (a_n) converges to p, then so does every subsequence. But then realized that (a_n) is ANY sequence, so my proof would not hold. I'm also not sure if a sequence even does converge to p!
Any ideas how to prove this result?