Given topological spaces $(X,\tau)$, $(Y,\upsilon)$, we say a function $f:X\rightarrow Y$ is continuous iff $\forall$ $U\in \upsilon$, $f^{-1}(U)\in \tau$.

Equivalently, we can say that $f$ is continuous iff $\forall$ convergent nets $x_i \rightarrow x$ in $X$, $f(x_i) \rightarrow f(x)$ in $Y$.

I understand that we need the generalization of sequences (nets) since not all topological spaces are first-countable.
Is there an example to illustrate the point? That is, is there a construction of a function $f$ s.t. $f$ is not continuous (using the first definition), but $f$ satisfies the second definition when limited to sequences? Just curious.

  • $\begingroup$ In this question there is a description of such a function. $\endgroup$ – t.b. Sep 13 '11 at 3:48
  • $\begingroup$ @Theo: That's quite the elaborate example. Nevertheless, thanks for the link. Can't hurt to see other constructions. $\endgroup$ – Dustin Tran Sep 13 '11 at 4:14
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    $\begingroup$ Yes, I agree it's much harder than Pete's example, but it is also much more interesting, I think. There you have a Hausdorff space and there are many non-constant sequences. There's just one point (called $b$ there) that can't be reached by a non-constant sequence, and that's why you can assign an arbitrary value to a function there without destroying sequential continuity. On the way of the construction you also get a space that is sequentially compact (every sequence has a convergent subsequence) but not compact. $\endgroup$ – t.b. Sep 13 '11 at 4:22

Let $X$ be an uncountable set endowed with the cocountable topology $\tau$: a nonempty subset is open iff its complement is countable. This is a non-discrete space in which the only convergent sequences are the eventually constant sequences.

It follows that for any topological space $Y$ and any function $f: X \rightarrow Y$, $f$ is sequentially continuous, i.e., if $x_n \rightarrow x$ then $f(x_n) \rightarrow f(x)$. But because $X$ is not discrete, the identity map from $(X,\tau)$ to $(X,\text{discrete})$ is not continuous.

  • $\begingroup$ Oh, I see now! Thanks for the quick response. I wasn't aware that there was even a term for such a thing. $\endgroup$ – Dustin Tran Sep 13 '11 at 3:57
  • $\begingroup$ @Dustin: you're welcome. There are terms for lots (maybe even most?) things in mathematics, but many of these terms are not widely known. For instance, I wrote a fairly substantial set of notes on convergence in topological spaces -- math.uga.edu/~pete/convergence.pdf -- and I just checked that the term "sequentially continuous" does not appear there. (But the above example does appear as a non-discrete space for which there are no nontrivial convergent sequences, which is why I quickly knew the answer to your question...) $\endgroup$ – Pete L. Clark Sep 13 '11 at 4:04
  • $\begingroup$ @Landyn: The same thing can be done for any space which is not sequential. If $(X,\mathcal T)$ is not sequential, take all sequentially open sets; they form a new topology $\mathcal T_s$ on $X$. The spaces $(X,\mathcal T)$ and $(X,\mathcal T_s)$ have the same convergent sequences and $\mathcal T\subsetneq \mathcal T_s$. So the identity map $id_X \colon (X,\mathcal T_s)\to(X,\mathcal T)$ is sequentially continuous but not continuous. For $\mathcal T$=cofinite you get $\mathcal T_s$=discrete. $\mathcal T_s$ is called sequential coreflection. $\endgroup$ – Martin Sleziak Apr 27 '12 at 6:20

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