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Topology and Sequences?
I'm having a real issue with this proof. I'll state the actual problem then show what I've got so far. Any help would be fantastic.
Prove the following: Let A is a subset of ℝⁿ with the standard topology. If x is a limit point of A, then there is a sequence of points in A that converges to x.
Attempted proof:
x being a limit point of A means that every neighborhood of x (open balls) intersects A at some point(s) other than x itself. As ℝⁿ is Hausdorff, every convergent sequence converges to a unique point.
And that's where I get stuck. It's kind of obvious that you could find a sequence, but I just can't think of how to say it or have whatever I say to be convincing.
Thanks for the help.
Proof for {x_n} -> x:
As n -> ∞, {x_n} -> x because x_n ∈ B₀(x,n) = {y | 0 < d(x,y) < 1/n} which converge to the center of the balls, x itself, because 1/n -> 0 as n -> ∞, this is well known.
I think that works.
2 Answers
- No MythologyLv 71 decade agoFavorite Answer
Construct it. Consider the deleted open balls of the form B_0(x, n) = {y | 0 < d(x,y) < 1/n}. As x is a limit point of A,
B_0(x,n) ∩ A ≠ Φ.
So, for each n, choose x_n ∈B_0(x,n) ∩ A. This sequence {x_n} ⊂ A, and all you have to do is argue that x_n -> x. This isn't hard to do because the radius of the sequence of balls is 1/n.
- Anonymous5 years ago
finite countable? ... isn't that the same thing as finite? Anyway, I don't think it's true in general, since the statement requires the discrete topology on X. Maybe I don't understand your statement, but T doesn't seem to necessarily be discrete. Maybe you mean "first countable" (don't know if that's enough either, but at least it would make more sense than "finite countable")?
