# convergence of nets vs. convergence of really long sequences

It is well known that for a function $f:X\to Y$ between the underlying sets of topological spaces, the condition that $f$ is continuous is equivalent to the condition that given any net $N$ in $X$ that converges to $x$, the net $f(N)$ converges to $f(x)$.

I would like to know if a similar characterization of continuity can be obtained using more restricted nets. In particular, is it true that $f$ is continuous provided that given any linear net $L$ in $X$ converging to $x$, the net $f(L)$ converges to $f(x)$.

I would also like to know if some upper bound on the size of the net can be placed (depending on the cardinalities of $X$ and $Y$ or the cardinalities of their topologies). In more detail, given $X$ and $Y$, is there some cardinal $\kappa$ such that $f$ will be continuous provided that given a net (resp. linear net) $N$ of cardinality smaller than $\kappa$ that converges to $x$, the net $f(N)$ converges to $f(x)$.

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In case anyone else was wondering, a linear net (according to this definition) is a net whose directed set is linearly ordered. – joriki Nov 8 '12 at 6:06
thanks joriki for the clarification. That is indeed what I meant. – Ittay Weiss Nov 8 '12 at 6:34
This is somewhat similar question which asks about nets on well-ordered sets: Is a net stronger than a transfinite sequence for characterizing topology?. The following paper studies spaces, where topology is determined by linearly ordered nets: James R. Boone: A note on linearly ordered net spaces. Pacific J. Math. Volume 98, Number 1 (1982), 25-35; link. – Martin Sleziak Nov 16 '12 at 5:28
Thanks Martin Sleziak – Ittay Weiss Nov 16 '12 at 6:12

Linear nets do not in general suffice.

Let $p$ be a free ultrafilter on $\omega$, and let $X=\{p\}\cup\omega$. Points of $\omega$ are isolated, and $U\subseteq X$ is a nbhd of $p$ iff $p\in U$ and $U\setminus\{p\}\in p$. A map $f:X\to X$ is continuous iff $f(p)=p$ and $f^{-1}[U]\in p$ for each $U\in p$.

Note that $p$ is not the limit of any sequence in $\omega$, so the only convergent sequences in $X$ are the trivial ones, those that are eventually constant. Now suppose that $\nu$ is a linear net in $\omega$ of uncountable cofinality. There must be some $n\in\omega$ that appears cofinally in $\nu$, which is therefore not eventually in the nbhd $X\setminus\{n\}$ of $p$. Thus, no linear net in $\omega$ converges to $p$, and the only convergent linear nets in $X$ are trivial. Since every function preserves the convergence of trivial nets, but not every bijection $f:X\to X$ preserving $p$ is continuous, linear nets do not suffice to characterize the continuous maps from $X$ to $X$.

In general let $\kappa=\sup\{\chi(x,X)^+:x\in X\}$, where $\chi(x,X)$ is the character of the point $x$ in $X$, i.e., the maximum of $\omega$ and the minimum cardinality of a local base at $x$. Then every point of $X$ has a local base of cardinality less than $\kappa$. Suppose that $f:X\to Y$ is not continuous. Then there are an $x\in X$ and an open nbhd $U$ of $f(x)$ in $Y$ such that for every open nbhd $V$ of $x$, $f[V]\nsubseteqq U$. Let $\mathscr{B}$ be a local base at $x$ of minimum cardinality. $\langle\mathscr{B}\supseteq\rangle$ is a directed set of cardinality less than $\kappa$. For each $B\in\mathscr{B}$ fix $x_B\in B\setminus f^{-1}[U]$; then $\langle x_B:B\in\mathscr{B}\rangle$ is a net in $X$ converging to $x$ such that $\langle f(x_B):B\in\mathscr{B}\rangle$ does not converge to $f(x)$, so $f$ is not continuous. Thus, if a function $f:X\to Y$ preserves limits of convergent nets of cardinality less than $\kappa$, $f$ is continuous.

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Thanks Brian for this really nice answer. For the first part, since you use a free ultrafilter, you use the axiom of choice. Do you know if AC is required ? – Ittay Weiss Nov 8 '12 at 8:47
@Ittay: You’re welcome. As for AC, I’ve no idea: AC is part of my standard equipment, and I’ve never had much interest in doing without it. (I do find the relationships amongst its various weaker consequences moderately interesting, but I just don’t care about them when I’m doing ‘everyday mathematics’, so to speak.) – Brian M. Scott Nov 8 '12 at 8:51
@Ittay: There are some papers called the horrors of topology without choice. It is indeed a horrific place. If you wish to talk about non-AC universes then topological spaces whose underlying sets cannot be well ordered are such that really long sequences are useless. – Asaf Karagila Nov 8 '12 at 8:59
Brian, just for the record, I was totally happy with the answer but was just curious as to the use of AC. Thanks again. Asaf, glad you joined in with this remark. – Ittay Weiss Nov 8 '12 at 9:02
@Ittay: If you’re curious about how bad things can get, I recommend starting with Eric van Douwen’s paper Horrors of topology without AC: a nonnormal orderable space: Eric was a better expositor than most, and it’s one of the earliest papers in that area. – Brian M. Scott Nov 8 '12 at 9:09