Talk:Space of mappings, topological

Topology of uniform convergence

The usage of the phrase topology of uniform convergence here does not seem to be the same as that of the article Topology of uniform convergence. Richard Pinch (talk) 17:54, 29 December 2016 (CET)

Yes! And moreover, this is probably an irrecoverable error of the author (Arkhangel'skii); the same error manifests itself in the Russian version. First, the topology of uniform convergence really depends on the uniform structure (not just topology) on Y. Second, even for Y=[0,1] I do not see how to define the topology of uniform convergence that way (no matter what is the family S) when it is different from the compact-open topology (which happens easily when X is not compact). Boris Tsirelson (talk) 22:04, 29 December 2016 (CET)

Indeed, consider $X=\R$ and $Y=[0,1]$. If $S$ contains (at least one) non-compact set $A$, then $\mathfrak{T}$ is too strong, since there exists a continuous $f:X\to Y$ such that $f[A]\subseteq(0,\infty)$ and $ \inf_{x\in A} f(x)=0 $; the set of all continuous $g:X\to Y$ such that $g[A]\subseteq(0,\infty)$ does not contain $f-\varepsilon$, thus, is not a neighborhood of $f$ in the topology of uniform convergence. And if $S$ contains only compact sets, then $\mathfrak{T}$ is too weak, weaker than (or equal to) the compact-open topology, that is (in this case) the topology of locally uniform convergence. And the phrase "if $X\in S$, then $\mathfrak{T}$ is called..." in the article makes no sense, since it does not specify, what else is included into $S$. I propose to remove all that from the article. Boris Tsirelson (talk) 07:48, 30 December 2016 (CET)