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Let $X$ and $Y$ be topological spaces. Then for any directed set $A$ we can define nets in $X$ and $Y$ indexed by $A$. And a function $f:X\rightarrow Y$ is continuous at a point $x$ if and only if all $f$ maps all nets converging to $x$, indexed by all directed sets, to nets converging to $f(x)$.

But something stronger than that is true. Let $A_0$ is the set of all open sets containing $x$, endowed with a partial order $\leq$ defined by saying that $U\leq V$ if $V$ is a subset of $U$. Then $f$ is continuous at $x$ if and only if $f$ maps all nets converging to $x$ indexed by $A_0$ to nets in $Y$ converging to $f(x)$. What that means is that if $f$ maps all nets converging to $x$ indexed by $A_0$ to nets converging to $f(x)$, then $f$ maps all nets converging to $x$ indexed by all directed sets to nets in converging to $f(x)$.

My question is, are there any directed sets other than $A_0$ which have this property? The natural numbers under the standard ordering clearly do not have this property, unless the topological space is well-behaved, since a function $f$ mapping sequences converging to $x$ to sequences converging to $f(x)$ can fail to be continuous at $x$. What about the real numbers, or the ordinal numbers?

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  • $\begingroup$ @WilliamElliot Thanks, I fixed it. $\endgroup$ – Keshav Srinivasan Sep 16 '18 at 3:01
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    $\begingroup$ It turns out you can do quite a lot with transfinite sequences (nets with an ordinal as the domain) instead of just arbitary nets. See this paper by Howes e.g. $\endgroup$ – Henno Brandsma Sep 16 '18 at 4:51

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