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Given a topological space $\mathcal{X}=(X,\tau)$ and a set $A\subseteq X$, say that $A$ is $\mathcal{X}$-sufficient iff every continuous function $(A,\tau_A)\rightarrow\mathbb{R}$ extends to a unique continuous function $\mathcal{X}\rightarrow\mathbb{R}$ (where we equip $\mathbb{R}$ with the usual topology). Let $\mathfrak{suff}(\mathcal{X})$ be the smallest cardinality of an $\mathcal{X}$-sufficient set; what is the general term for this (and what are some good sources)?

Note that we can replace $\mathbb{R}$ with any topological space $\mathcal{C}$ here and get the analogous notions $(\mathcal{X},\mathcal{C})$-sufficiency and and $\mathfrak{suff}_\mathcal{C}(\mathcal{X})$. Changing the target space, even staying in the realm of "rich" target spaces, can drastically alter the value of the relevant sufficiency cardinal. For example, $\mathfrak{suff}(\mathbb{R})=2^{\aleph_0}$, since every disconnected subspace admits a continuous function to $\mathbb{R}$ not extendible to all of $\mathbb{R}$ and every subspace missing an interval has too many continuous extensions of (say) the constant-zero function (so the only sufficient subspace is $\mathbb{R}$ itself), but $\mathfrak{suff}_\mathcal{\mathcal{D}}(\mathbb{R})=1$ for every totally disconnected $\mathcal{D}$. I'm more broadly interested in the whole function $\mathfrak{suff}_{-}(-)$, but the specific case $\mathcal{C}=\mathbb{R}$ seems more likely to have a lot of material easily available.

(This is related to a variation of this recent question of mine - basically, given a first-order sentence $\varphi$ and a space $\mathcal{X}$ with $C(\mathcal{X},\mathbb{R})\models\varphi$, how small a subspace $\mathcal{Y}\subseteq\mathcal{X}$ can I find such that $C(\mathcal{Y},\mathbb{R})\models\varphi$?)

Noah Schweber
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  • Interesting question (can't add anything helpful). There might be a word missing here: "since any proper subspace of R is disconnected and ..." – Randall Aug 02 '21 at 17:27
  • @Randall Nope, that's right as written (I did almost type "write as righten" just now, though :P): "and so" could be replaced with "and therefore" or etc. – Noah Schweber Aug 02 '21 at 18:15
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    @AlessandroCodenotti $$x\mapsto\begin{cases}0 &\mbox{if }x<\pi\ 1 & \mbox{ if }x>\pi.\\end{cases}$$ (That said, I did write that every proper subspace of $\mathbb{R}$ is disconnected, which ... is bonkers.) – Noah Schweber Aug 02 '21 at 22:43
  • Ah of course, I want uniform continuity to extend from a dense subset – Alessandro Codenotti Aug 02 '21 at 22:45
  • Actually isn't $A=\Bbb R$ the only subspace of $\Bbb R$ that works? That seems reasonable since for completely regular spaces the ring of continuous real valued functions determines the space – Alessandro Codenotti Aug 02 '21 at 23:02
  • @AlessandroCodenotti Yes, exactly: a proper subspace either has a cut point or misses an interval. I've edited to make this explicit. – Noah Schweber Aug 02 '21 at 23:11
  • Potentially relevant: https://mathoverflow.net/questions/136554/generalizations-of-the-tietze-extension-theorem-and-lusins-theorem – HallaSurvivor Aug 03 '21 at 00:19
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    If $X$ has a point $x\in X$ such that $x$ is a limit of some non-constant sequence and $U=X\backslash{x}$ is normal then this $suff(X)=|X|$. That's because we take a sequence $(x_n)$ convergent to $x$ inside $U$. This sequence is closed in $U$ and thus by Tietze any function $f:{x_n}\to\mathbb{R}$ extends to $U$. We can choose $f$ so that $f(x_n)$ diverges. In that case the extension $F:U\to\mathbb{R}$ cannot be extended further to full $X$. In particular this makes all hereditary normal, first countable spaces (for example all metrizable spaces) trivial, right? – freakish Aug 05 '21 at 07:58

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