Let $X=\omega +1$ be convergent sequence. Then what does mean by "$X$ is convergent sequence"?
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1$\begingroup$ Where does this sentence appear? I would guess it means that a sequence $\omega \to Y$ is convergent if and only if it can be extended to a map $X \to Y$ that is continuous for the coinitial topology on $X$, but it's impossible to say without context. $\endgroup$– LSpiceCommented Aug 15, 2022 at 19:09
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$\begingroup$ In view of @AndrejBauer's answer, I hope this does not get deleted. $\endgroup$– Steven LandsburgCommented Aug 16, 2022 at 2:27
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$\begingroup$ I agree the question needs clarity, but it need not be closed because the answer instills all the clarity required. $\endgroup$– Andrej BauerCommented Aug 16, 2022 at 7:25
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2$\begingroup$ @AndrejBauer: I believe that MSE would have been a much more appropriate home for this question than MO. Voting to keep closed. (Oh, and the OP is one of those Stack Exchange users who never accept answers.) $\endgroup$– Alex M.Commented Aug 16, 2022 at 7:46
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$\begingroup$ Oh well, let's say it's service to the community then :-) $\endgroup$– Andrej BauerCommented Aug 16, 2022 at 8:15
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1 Answer
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Consider $\omega + 1 = \{0, 1, 2, \ldots, \omega\}$ with the interval topology. Then every finite $n < \omega$ is an isolated point, and the sequence $0, 1, 2, 3, \ldots$ converges to $\omega$.
Given a space $X$, let $\mathsf{Conv}(X) = \{ (a, x) \in X^\omega \times X \mid \lim_n a_n = x\}$ be the space of convergent sequences with their limits. Observe that $\mathsf{Conv}(X) \cong \mathcal{C}(\omega + 1, X)$, i.e., convergent sequences with limits in $X$ are the same thing as continuous maps $\omega + 1 \to X$. This is why the $\omega + 1$ with the interval topology deserves to be called the generic convergent sequence. Every convergent sequence is its continuous image, and every continuous image is a convergent sequence.
answered Aug 15, 2022 at 19:38