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Theorem

Let $X$ and $Y$ two topological spaces: if $X$ is first countable and if $f:A\rightarrow Y$ is a continuous function such that $f(x)=y_0$ for any $x\notin S$ where $S$ is a closed set contained in $\overset{\,\,\circ}{A}$ then the function $F:X\rightarrow Y$ defined through the condiction $$ F(x):=\begin{cases}f(x_0)\,\,\text{if}\,x_0\in A\\y_0,\,\,\text{otherwise}\end{cases} $$

is a continuous extension of $f$ to $X$.

Before to prove this theorem we remember some fundamental results.

Lemma 1

If $X$ is first countable and if $f:X\to Y$ is a function then $y_0$ is the limit of $f$ as $x$ approaches at $x_0$ if and only if for any sequence $(x_n)_{n\in\Bbb N}$ converging to $x_0$ it happens that $\left(f(x_n)\right)_{n\in\Bbb N}$ converges to $y_0$.

Proof. See here.

Lemma 2

If $X$ is first countable and if $f:X\rightarrow Y$ is a continuous function then any sequence $(x_n)_{n\in\Bbb N}$ converges to $x_0$ if and only if the sequence $(f(x_n))_{n\in\Bbb N}$ converges to $f(x)$.

Proof. See page $71$ of "General Topology" by Stephen Willard.

Theorem 3

If $X$ is first countable and if $f:X\rightarrow Y$ if a function then it is continuous if and only if $f(x_0)$ is the limit of $f$ as $x$ approaches at $x_0$.

Proof. A simple consequence of the two preceding lemmas.

Lemma 4

If $(x_n)_{n\in\Bbb N}$ is a sequence converging to $x_0$ and if $(y_n)_{n\in\Bbb N}$ is a sequence such that $y_n=x_n$ for any $n\ge n_y$ then $(y_n)_{n\in\Bbb N}$ converges to $x_0$ too.

Proof.If $y_n=x_n$ for any $n\ge n_y$ if for any neighborhood $U$ of $x_0$ there exist $n_U$ such that $x_n\in U$ for any $n\ge n_U$ then $y_n\in U$ for any $n\ge\max\{n_y,n_U\}$ so that the lemma holds.

So now we prove the theorem:

Proof. To prove the theorem we use the theorem 3 and in particular we separately analyze the case where $x_0\in\overset{\,\,\circ}A$, the case where $x_0\in\partial A$ and the case where $x_0\in\text{ext}(A)$ since the collection $\mathcal{P}:=\{\overset{\,\,\circ}A,\,\partial A,\text{ext}(A)\}$ is a partition of $X$.

  • So if $x_0\in\text{ext}(A)$ then the sequence $(F(x_n))_{n\in\Bbb n}$ is semicostant (or rahter $F(x_n)=y_0$ for any $n\ge n_0$) for any sequence $(x_n)_{n\in\Bbb N}$ converging to $x_0$ so that we conclude that the function $F$ is continuous in $\text{ext}(A)$.

  • Since $S=\overline S\subseteq\overset{\,\,\circ}A$ then if $\overline S\cap\partial A=\emptyset$ so that for any $x_0\in\partial A$ there exist a neighborhood $U$ such that $(U\cap S)=\emptyset$ and so if $(x_n)_{n\in\Bbb n}$ is a sequence converging to $x_0$ there exist $n_0\in\Bbb N$ such that $x_n\in U$ for any $n\ge n_0$ that is $F(x_n)=y_0$ for any $n\ge n_0$ so that as above we conclude that $F$ is continuous in $\partial A$.

  • Finally if $(x_n)_{n\in\Bbb n}$ is a sequence converging to $x_0\in\overset{\,\,\circ}A$ then there exist $n_0\in\Bbb N$ such that $F(x_n)=f(x_n)$ for any $n\ge n_0$ and so by lemma $4$ we conclude that $F$ is continuous in $\overset{\,\,\circ}A$ too.

So the theorem holds.

Although the proof of the theorem seems correct I doubt that it is false. Indeed if $X=Y=\Bbb R$ and $f(x):=sin(\frac{1}x)$ then $A:=\Bbb R\setminus\{0\}$ and $S=\{\frac{1}{k\pi}:k\in\Bbb Z\setminus\{0\}\}$ then $$ F(x):=\begin{cases}sin(\frac{1}x)\,\,\text{if}\,x\neq 0\\0,\,\,\text{otherwise}\end{cases} $$

and this function is not continuous: any way I am not sure that $S$ is closed because it seems to me that $0\in\overline S$ and $0\notin S$. So is the theorem true? and if yes, is my proof correct? and if the theorem false is my counterexample correct? So could someone help me, please?

VividD
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Antonio Maria Di Mauro
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  • Your counter-example is false at least: $S$ isn’t closed, as $0 \notin S$ but $0 \in \overline{S}$. – Aphelli Aug 07 '20 at 11:18
  • @Mindlack Okay, and what can you say about the truthfulness of the theorem? – Antonio Maria Di Mauro Aug 07 '20 at 11:19
  • $F$ restricted to the interior of $A$ is continuous, $F$ restricted to $X \backslash S$ is continuous (constant), so as both are open subsets of $X$ and since they cover $X$, $F$ is continuous. So your theorem is right. – Aphelli Aug 07 '20 at 11:22
  • @Mindlack Okay, but you don't consider the case where $x_0\in\partial A$. Anyway what can you say about my proof? – Antonio Maria Di Mauro Aug 07 '20 at 11:25
  • I covered all cases, because $\partial A \subset X \backslash S$. Anyway, your proof looks correct (the first case could do with a little more explanation, to be in line with the two others though). – Aphelli Aug 07 '20 at 11:30
  • @Mindlack Okay, thanks too much for your assistance. – Antonio Maria Di Mauro Aug 07 '20 at 11:31
  • Or is there one fixed $S$ you're talking about? Then you just need $f \equiv y_0$ on $\partial A$ and apply the pasting lemma as I did recently here.. – Henno Brandsma Aug 07 '20 at 11:41
  • @HennoBrandsma So $S$ is a closed set in $X$ and if $x\in A\setminus S$ then $f(x)=y_0$. – Antonio Maria Di Mauro Aug 07 '20 at 11:44
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    This implies $f \equiv y_0$ on $\partial S$ and the previous pasting argument applies. – Henno Brandsma Aug 07 '20 at 11:45
  • @HennoBrandsma Okay, so if I understood the pasting lemma says that if $f:A\rightarrow Y$ and $g:\overline{X\setminus A}\rightarrow Y$ are continuous then the function $f\cup g$ is continuous too, right? – Antonio Maria Di Mauro Aug 07 '20 at 11:50
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    No it says that if $X=A \cup B$ and $f: A\to Y,g:B \to Y$ are continuous and $f$ restricted to $A \cap B$ equals $g$ restricted to $A \cap B$ then the combined function on $X$ (which is then well-defined) is continuous too when $A$ and $B$ are both closed (or both open). – Henno Brandsma Aug 07 '20 at 11:54
  • @HennoBrandsma Okay, but unfortunately in my case I don't know if $A$ is closed or open: so how apply the pasting lemma in the above case? – Antonio Maria Di Mauro Aug 07 '20 at 12:39
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    $S$ is closed and so is $\overline{S^\complement}$. – Henno Brandsma Aug 07 '20 at 12:40
  • Okay, so $S\cap\overline{S^C}=\partial S$ thus I have to observe that $f(x)=\lim_{x\rightarrow x_0}f(x)=y_0$ for $x_0\in\partial S$ (indeed $S\subseteq A$ and the function $f$ is continuous in $A$) and that the function $F$ is the combined function on $X$ of the restriction to $S$ of the function $f$ defined in $A$ and the costant function $y_0$ defined in $\overline{S^C}$, right? – Antonio Maria Di Mauro Aug 07 '20 at 12:48
  • Limits are not needed, nor is consideration of the boundary.. See my answer. – Henno Brandsma Aug 07 '20 at 12:56

1 Answers1

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There is no need for sequences nor first countability: your $F$ satisfies $F\restriction_S = f$ which is continuous and $F\restriction_B \equiv y_0$ where $B = \overline{S^\complement}$ so also continuous so the pasting lemma tells us that (as $S \cup B=X$ and $S,B$ are both closed) that $F$ is continuous. You could use the partition into two open sets $A^\circ$ and $S^\complement$ as well.

Henno Brandsma
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  • Okay, now it is all clear. Thanks too much for your assistance!!! – Antonio Maria Di Mauro Aug 07 '20 at 12:59
  • @AntonioMariaDiMauro You're welcome. I hope this statement did not come from a book ? It has unneccesary assumptions. – Henno Brandsma Aug 07 '20 at 13:01
  • The statement if a my conjecture: I did it to explain the continuity of the extension of the function $f:\text{int}J\rightarrow\Bbb R$ defined in the Munkres's proof (see page 163 of "Analysis on Manifold" by James Munkres) of the change of variables theorem. – Antonio Maria Di Mauro Aug 07 '20 at 13:14
  • @AntonioMariaDiMauro I don't have that book by Munkres, just the first edition of his topology text. – Henno Brandsma Aug 07 '20 at 13:19
  • If you like you can download it here. – Antonio Maria Di Mauro Aug 07 '20 at 13:19
  • Precisely in the case of the change of variables theorem the set $S$ is the support of $f$ that by definition is closed and contained in $\text{int},J$. – Antonio Maria Di Mauro Aug 07 '20 at 13:21
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    @AntonioMariaDiMauro If he'd assumed his topology text as a prerequisite, he could have used his pasting lemma.. – Henno Brandsma Aug 07 '20 at 13:28
  • I believe that it do this: although in the first chapter (The Algebra and Topology in $\Bbb R^n$) he show some fundamental result of topology he implicitly assumes that the reader studied Topology so the first chapter is only a summary of things that the reader knows. – Antonio Maria Di Mauro Aug 07 '20 at 13:32
  • This is the reason that induced me to study first Set Theory and Topology before to study Differential Geometry: indeed with support of these two things many passages of Differential Geometry's proofs can be explained more simply. – Antonio Maria Di Mauro Aug 07 '20 at 13:39
  • Excuse me but I have doubt: I don't understand why $F|_B\equiv y_0$ because if $x\in\partial S$ then $F|_B(x)=F(x):=f(x)$ but I don't know if in this case it is $f(x)=y_0$. Could you explain this, please? – Antonio Maria Di Mauro Aug 08 '20 at 11:39
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    @AntonioMariaDiMauro Your assumptions imply that $f$ is always $y_0$ in a boundary point of $S$. Just note that $\partial S \subseteq \overline{S^\complement}$ and that $$f [ \overline{S^\complement} ] \subseteq \overline{ f[S^\complement]} = \overline{{y_0}}= {y_0}$$ and hence $f[\partial S]={y_0}$. So I do assume $X$ is $T_1$, which is reasonable in light of the origin of the idea. – Henno Brandsma Aug 08 '20 at 13:54
  • Okay, I understand: however if to prove the continuity of $F$ we use the open set $\overset{,,,\circ}A$ and $S^C$ is not necessary to suppose that $X$ is $T_1$, right? – Antonio Maria Di Mauro Aug 08 '20 at 18:39