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Let $(X,d)$ be a metric space, denote by $\tau_d$ the induced metric topology. Let $G_i$, $i\in I$, be a countable union of disjoint $G_\delta$-subsets. Mind that it could be that $G_i \notin \tau_d$ and $\cup_{i \in I}G_i$ could be a proper subset of $X$. Consider continuous maps $f_i:(G_i,d_{G_i}) \mapsto (Y,h)$, where $d_{G_i}$ is the restriction of $d$ to $G_i$ and $(G,h)$ is a metric space.

Question is it possible to patch together the maps $f_i$'s and extend the map so obtained, say $f$, to a continuous map $f:(X,d)\mapsto (Y,h)$?

Remark 1 Assume that in particular $G_i$'s are open and cover $X$, i.e. $X=\cup_{i \in I}G_i$. Then the disjoint union topological space $\coprod_{i\in I}(G_I,d_{G_i})$ and $(X,d)$ should coincide (since each ${\tau_{d_{G_i}}}$ would consists of intersections between $G_i$ and open sets, say $U$, of $\tau_d$; but then also $G_i\cap U \in \tau_d$). Then $f=f'$ would exists by the universal property of coproducts.

Remark 2 Let's now give up the assumption $X=\cup_{i \in I}G_i$, but retain $G_i \in \tau_d, \, \forall i \in I$. Endow $G=\cup_{i \in I}G_i$ with the restriction of $d$, say $d_G$. Then, we still have $G_i \in \tau_G$ and $f$ obtains as in the previous remark, but now as a continuous map from $(G,d_G)$ to $(Y,h)$. If $G$ is dense in $X$, then there exists also a continuous map $f'$ extending $f$ to $X$ (but it may not be unique, see e.g. the answer to this question Extending a function by continuity from a dense subset of a space ).

Remark 3 Let's finally give up also the assumption $G_i \in \tau_d, \, \forall i \in I$. If there existed $\tau_d$-open, disjoint sets $G_i'$ such that $G_i=G \cap G_i'$, then the $G_i \in \tau_{d_G}$, then it should be possible to obtain $f:(G,d_G)\mapsto (Y,h)$ as above and, if $G$ is again dense, then we can extend $f$ to some continuous $f'$.

Does the third remark make sense? Is it always possible to construct the aforementioned open sets $G_i'$? (If yes, this is not immediately clear to me). Can we obtain a positive answer to the above main question also beyond the restrictions in Remark 3? (That was the more general setting I managed to handle so far). Any help or refs are appreciated.

Jack London
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Every singleton set in $\mathbb R$ is a $G_{\delta}$ set. Consider the collection of sets $\{\{r\}: r \in \mathbb Q\}$. This is a disjoint countable collection of $G_{\delta}$ sets. Let $f$ be any map from $\mathbb Q$ to $\mathbb R$. This gives a family of continuous functions on these singletons. Your question asks if we can find a continuous map extending these functions. The answer is obviously NO.

Kavi Rama Murthy
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  • Ok. But do you think that, at least in some cases, the sets $G_i'$ might exist and Remark 3 work well, or do you think it's incorrect? – Jack London Dec 10 '19 at 12:36
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The minimal requirement is of course that $f_i\mid_{G_i \cap G_j} = f_j\mid_{G_i \cap G_j}$ for all $i, j$. So let us assume this in the sequel.

Remark 2: It is false. Extending a function by continuity from a dense subset of a space deals only with extending uniformly continuous maps. Take for example $X = [0,1], G = (0,1]$ and $f : G \to \mathbb [0,1], f(x) = \sin(1/x)$. This has no continuous extension to $[0,1]$ since $\lim_{x \to 0}f(x)$ does not exist.

Remark 3: In that case the $G_i$ are open in $G$, thus you get $f : G \to H$ as desired. But there is no reason why an extension to $X$ should be possible.

Paul Frost
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  • Actually the answer to the linked question starts with ``If you only need continuity of $g$, then continuity (not uniform) of $f$ is enough", where in my case $g$ would be $f'$. Is the cited statement false? According to that answer, it would seem that uniformity is only needed for unicity. Or, more informally, if the extension is not univocally defined (for which uniform continuity is needed), is the extension substantially useless, since it is not well defined? – Jack London Dec 10 '19 at 15:13
  • In addition, in Remark 2, estension of $f$ to some $f'$ should be allowed by adding assumptions on $(Y,h)$, e.g. Theorem 1 A. in blair (1976) https://www.jstor.org/stable/2040816?seq=1#metadata_info_tab_contents – Jack London Dec 10 '19 at 15:31
  • It is definitely wrong that any continuous $f : D \to Y$ defined on a dense subset of a metric space $X$ has a continuous extension $F : X \to Y$. See my example for Remark 2. If $X$ is compact and $Y$ is complete, then $F$ exists iff $f$ is uniformly continuous. The acknowledged answer to the linked question does not define an extension $g$ of $f$, but only proves that the definition $g(x) = \lim_{x_n \to x, x_n \in D} f(x_n)$, if this limit always exists, yields a continuous $g$.See Srivatsan's answer to the question. – Paul Frost Dec 10 '19 at 16:28
  • Uniform continuity is irrelevant for the uniqueness of continuous extensions. In fact, any two contiuous maps $F : F' : X \to Y$ into a Hausdorff space $Y$ which agree on a dense subset $D$ agree on $\bar D = X$. See https://math.stackexchange.com/q/305472 – Paul Frost Dec 10 '19 at 16:33
  • Blair's paper gives conditions on $Y$ and on $f$ which guarantee the existence of an extension. It does not say that any $f$ has an extension. – Paul Frost Dec 10 '19 at 16:45
  • Sure. Anyway in Remark 2 at least a continuous map $f:(G,d_G)\mapsto (Y,h)$ obtains, right? So, if $G$ is dense and $(Y,h)$ is compact, the continuous extension to $X$ exists, correct? – Jack London Dec 10 '19 at 17:06
  • Yes, $f : G \to Y$ exists. But see my example where we have a single $G_1 = (0,1]$ (i.e. $G = G_1$). $X = Y = [0,1]$ are compact, but $f$ does not have a continuous extension to $X$. – Paul Frost Dec 10 '19 at 17:32
  • Ok, many thanks, I misinterpreted the reference by Blair. – Jack London Dec 11 '19 at 12:13