# on continuous function

Let the space $X=Y\cup\{a\}$, where $Y$ is a metriable subspace and $a \in \overline Y$. As we know, if $f$ is the coutinuous function from $Y^2$ to $I$, then $f$ can be continuously extendable over $X^2$. Since $Y$ is a metrizable space, there exist some continuous functions from $Y^2$ to $I$ satisfying $f(y_1,y_2)\neq 0$ whenever $y_1 \neq y_2$; $f(y_1,y_2)=0$ whenever $y_1 = y_2$. Therefore, the extensions $f^\sim (a,a)=0$. However, I don't know whether there exists one of those extensions satisfying that $f^\sim (a,y)\neq0$ and $f^\sim (y,a)\neq0$ for any $y \in Y$. Could someone help me?

-
So $X$ is not metrizable, and $Y$ is, I gather? Is it given that we can always extend functions from $Y^2$ to $I$, to $X^2$, or is it merely a possibility you are considering? In general there will be no extension. Note that given $f$, if there is an extension, it will be unique.... – Henno Brandsma Dec 21 '11 at 12:39
It seems that if such an extension exists, $a$ must be a $G_\delta$ in $X$: defining $h(x) = \overline{f}(x,a)$ for such an extension, $\{a\} = h^{-1}[\{0\}]$, which is a $G_\delta$. – Henno Brandsma Dec 21 '11 at 13:36
Idea for a counterexample: try the reals in the discrete topology and for some (suitable?) free ultrafilter $F$ on that set let $a$ be a new point with neighbourhoods $\{a\} \cup A$ for $A \in F$. – Henno Brandsma Dec 21 '11 at 13:43
$Y^2$ is dense in $X^2$, so the extension always exists! – Paul Dec 21 '11 at 13:50
this is clearly false, as $\sin(1/x)$ is well defined on $\mathbb{R}\setminus\{0\}$ but has no extension to $\mathbb{R}$ even though the former set is dense in the latter. Being able to continuously extend real-valued functions is quite special, and has a whole theory related to compactifications and completions etc. – Henno Brandsma Dec 21 '11 at 13:59

Let (C) denote the condition that a function $f$ defined on $Y\times Y$ is such that $f(x,y)\ne0$ for every $x\ne y$ in $Y$, $f(y,y)=0$ for every $y$ in $Y$.

Here is a positive example: assume that $X$ is metrizable, let $g(x,y)=d(x,y)$ for every $x$ and $y$ in $X$, and let $f$ denote the restriction of $g$ to $Y\times Y$. Then $f^\sim=g$ is the unique continuous extension of $f$ to $X\times X$, $f$ fulfills condition (C) and $f^{\sim}(a,y)=f^{\sim}(y,a)\ne0$ for every $y$ in $Y$.

Here is a negative example: assume that $X$ is metrizable, let $g(x,y)=d(x,y)d(a,x)d(a,y)$ for every $x$ and $y$ in $X$, and let $f$ denote the restriction of $g$ to $Y\times Y$. Then $f^\sim=g$ is the unique continuous extension of $f$ to $X\times X$, $f$ fulfills condition (C) and $f^{\sim}(a,y)=f^{\sim}(y,a)=0$ for every $y$ in $Y$.

These two examples show that some continuous functions $f$ on $Y\times Y$ such that condition (C) holds have a continuous extension $f^\sim$ to $X\times X$ such that $f^{\sim}(a,y)=f^{\sim}(y,a)\ne0$ for every $y$ in $Y$ and that some others have not.

-
Didier Piau, $X$ is not a metriable space, and hence you can't define $d(x,y)$ on $X$. – Paul Dec 21 '11 at 11:46
Sure, but since, by the examples above, the condition you are interested in fails already in the metrizable case, it cannot hold in general. – Did Dec 21 '11 at 14:12

For all $y\in Y$, you will define $f^\sim (a,y) = \lim_{n \rightarrow\infty} f(y_n, y)$ for any sequence $y_n \mathop\longrightarrow\limits_{n\rightarrow\infty} a$ in Y.

As $f$ is the distance on Y, if $\lim_{n \rightarrow\infty} f(y_n, y) = 0$ then $y_n \mathop\longrightarrow\limits_{n\rightarrow\infty} y$. By uniqueness of the limit, then $y = a$ and $a \in Y$, a contradiction.

-
Why would there be any sequence converging to $a$? It seems that $X$ need not be metrizable or even first countable. – Henno Brandsma Dec 21 '11 at 13:34
Using ultrafilters we can find examples where no such sequence exists.... – Henno Brandsma Dec 21 '11 at 13:44
Oh God. This is terrible. So I just proved that if there is a sequence in $Y$ converging to $a$, it’s ok... – Elvis Dec 21 '11 at 14:30
Not yet, because different sequences might have different limits under $f$; we need something like uniform continuity on $Y^2$, I think plus the existence of sequences. – Henno Brandsma Dec 21 '11 at 14:37