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I been reading for several hours and not yet found a question put in this way. Given any topological space:

  1. Does every sequance in $X$ determine a countable subset of $X$?

  2. Do the sets that belong the a topology on $X$ (the open sets) separete these sequences (given 1. is true) and therefore supply a structure for "the set $X$" turning it into a space?

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2 Answers 2

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After a sequence of comments on the other answer, I believe Question 2 is asking:

Is a topology uniquely determined by its convergent sequences?

The answer to this question is no. For a counterexample, let $\omega_1$ be the first uncountable ordinal, and consider $\omega_1 + 1 = \omega_1 \cup \{\omega_1\}$ under the order topology. Every neighborhood of $\omega_1$ then contains infinitely many elements, thus $\omega_1$ is not an isolated point, but no sequence converges to $\omega_1$ (any countable increasing sequence of ordinals necessarily has a countable limit ordinal).

Sometimes we generalize the concept of sequences to nets, which do capture all information about the topology of a space.

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Question 1 No, the sequence $\{x_n=1\}_{n=1}^\infty$ in $\mathbb{R}$ is an infinite sequence finite subset of $X$. Unless you mean something else by "determine a countable infinite subset".

Question 2 I'm not entirely sure what you are asking here. You're question seems somewhat circular or trivial since the open sets of $X$ will by definition "supply a structure for "the set $X$" and turn it into a space" - the open sets form the topology of $X$. Moreover, it seems that your second question does not rely on whether or not Question 1 was true or false.

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I confused myself..lets just say a countable subset. –  Johan Jul 26 '13 at 14:06
    
@Johan, what is the definition of countable to you? Countable means bijection with $\mathbb{N}$? –  Sigur Jul 26 '13 at 14:14
    
@Johan Sequences are inherently countable (unless you are talking of some sort of transfinite sequence over the ordinals, however, I don't believe that is what you mean. –  Christian Bueno Jul 26 '13 at 14:15
    
What Im asking in 2 is if these sequances determine subsets then the whole purpose of the topology is to seperate them by open sets –  Johan Jul 26 '13 at 14:17
    
@Johan In what sense do you mean that a sequence "determines" a subset? A sequence in $X$ is a function $f:\mathbb{N}\to X$, and I interpreted the "determined" subset to be the image of $f$. Is this what you mean? –  Christian Bueno Jul 26 '13 at 14:21

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