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I know that if a given function $f$ between two topological spaces is continuous then the image of a convergent sequence is a convergent sequence, and $f$ preserves the limit in the sense that $x_n \to x \implies f(x_n) \to f(x)$.

The converse is true if the domain is first countable: If $X$ satisfies the first axiom of countability and $x_n \to x \implies f(x_n) \to f(x)$ for any convergent sequence in $X$ then $f$ is continuous.

Thus this still holds if the condition on the domain is dropped? I've been trying to find an counterexample using non first countable spaces like the Sorgenfrey's Line and real numbers with the cofinite topology as the domain but every time I try to break the continuity of the function I end up with convergent sequences whose images don't converge. I was wondering if the statement isn't actually true and can be proved without the first axiom using a different technique maybe.

deabo
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    This may answer your question, although I don't know much about the order topology. – Mark Schultz-Wu Jul 25 '17 at 18:30
  • I'm not familiar with ordinal numbers and those $\omega_1+1$ they're talking about, but this seems to cover my question indeed. I'll try to read about ordinals and understand what has been done there, thank you! – deabo Jul 25 '17 at 18:37
  • In general, looking for "sequential continuity not implying continuous" is probably a search phrase if you want to try to find another counterexample. – Mark Schultz-Wu Jul 25 '17 at 18:39
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    Take an uncountable space with the co-countable topology (the open sets are $\varnothing$ and those sets whose complement is at most countable). Then every convergent sequence is eventually constant, so every map from that to a topological space is sequentially continuous. But of course there are lots of discontinuous maps. – Daniel Fischer Jul 25 '17 at 18:47
  • @DanielFischer I see! This result about convergent sequences being eventually constant doesn't hold in the cofinite topology though, does it? If I had tried the co-countable instead maybe I'd have figured it out before. – deabo Jul 25 '17 at 18:55
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    Yes, in the cofinite topology, a sequence with all terms distinct converges to every point of the space. So it's too easy for a sequence to converge in the cofinite topology. Having only few convergent sequences makes it easier for a discontinuous map to be sequentially continuous. – Daniel Fischer Jul 25 '17 at 19:00
  • Those make very elegant and simple examples, thank you! – deabo Jul 25 '17 at 19:03
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    The Sorgenfrey line is first countable, so no counterexample there.. – Henno Brandsma Jul 25 '17 at 19:13
  • See this answer for a proof of properties of the co-countable topology. – Henno Brandsma Jul 25 '17 at 19:42

2 Answers2

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Let $\tau$ be the co-countable topology in $\mathbb R$. Then a sequence $x_n$ in $(\mathbb R, \tau)$ converges to $x$ if and only if $x_n$ is eventually constant. In fact, suppose the sequence is often different from $x$, that is, given any $n \in \mathbb N$ there is a $m \geq n$ such that $x_m \neq x$. Then the set $\mathbb R \setminus \{x_m; x_m \neq x\}$ is an open neighborhood of $x$ which doens't contain any element of sequence $x_n$, so $x_n$ can not converge to $x$.

This implies every map $f$ on $(\mathbb R, \tau)$ has the property $x_n \to x \implies f(x_n) \to f(x)$, even the ones which are discontinuous. In particular, $f$ from $(\mathbb R, \tau)$ onto the real numbers with the usual topology maping $x$ to $x$ has this property and is not continuous, since $f^{-1}((0,1))$ is not an open of $(\mathbb R, \tau)$.

deabo
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Let $X=\beta \mathbb N$ \ $\mathbb N$ where $\beta \mathbb N$ is the Cech-Stone (maximal) compactification of the discrete space $\mathbb N.$ Then $X$ is an infinite compact Hausdorff space with no isolated points, and with the property that the only convergent sequences in $X$ are "eventually-constant" sequences. In fact, if $S$ is an infinite subset of $X$ then the cardinal of $\overline S$ is $2^{2^{\aleph_0}}.$ So any $f:X\to X$ preserves convergent sequences, whether $f$ is continuous or not. Of course $X$ is not $1$st-countable.

DanielWainfleet
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