Extending a function by continuity from a dense subset of a space

Question

I am given two spaces $X$ and $Y$, both Hausdorff. I have defined a uniformly continuous function on a dense set $D$ of $X$ that goes to $Y$. So once you have defined a function on a dense subset, if you would like to extend this function to the whole set in a continuous manner, then there is at most one way to do this. For the extension to be sequentially continuous, all decisions are "already made" by determining the value of the function on the dense subset. The value of $f(x)$ simply must be $\lim f(x_{n})$ where $x_{n}$ are in $D$ and converge to $x$. I proved well defined and unique.

I am having trouble proving this is continuous, any help is appreciated.

Thanks.. And this is embarrassing, but can someone tell me how to accept an answer. Sorry...

Answer 1

In the light of Alex Youcis's comment, I will assume that $X$ and $Y$ are metric spaces. The following approach has been taken verbatim (apart from notation changes) from John Erdman, A ProblemText in Advanced Calculus (Chapter 24, pp. 146-147).

1. If $f : X \to Y$ is a uniformly continuous map between two metric spaces and $(x_n)$ is a Cauchy sequence in $X$, then $f(x_n)$ is a Cauchy sequence in $Y$.

2. Let $X$ and $Y$ be metric spaces, $S \subseteq X$, and $f : S \to Y$ be uniformly continuous. If two sequences $(x_n)$ and $(y_n)$ in $S$ converge to the same limit in $X$ and if the sequence $f(x_n)$ converges, then the sequence $f(y_n)$ converges and $\lim f(x_n) = \lim f(y_n)$.

Proof Hint. Consider the "interlaced" sequence $(x_1, y_1, x_2, y_2, x_3, y_3, \ldots)$.

Now, the main theorem.

Theorem. Let $X$ and $Y$ be metric spaces, $S$ a subset of $X$, and $f : S \to Y$. If $f$ is uniformly continuous and $Y$ is complete, then there exists a unique continuous extension of $f$ to $\overline S$. Furthermore, this extension is uniformly continuous.

Proof Hint. Define $g : \overline S \to Y$ by $g(a) = \lim f(x_n)$ where $(x_n)$ is a sequence in $S$ converging to $a$. First show that $g$ is well defined. To this end you must show that

• $\lim f(x_n)$ does exist, and
• the value assigned to $g$ at $a$ does not depend on the particular sequence $(x_n)$ chosen. That is, if $x_n \to a$ and $y_n \to a$, then $\lim f(x_n) = \lim f(y_n)$.

Next show that $g$ is an extension of $f$.

To establish the uniform continuity of $g$, let $a$ and $b$ be points in $\overline S$. If $(x_n)$ is a sequence in $S$ converging to $a$, then $f(x_n) \to g(a)$. This implies that both $d(x_j , a)$ and $d (f(x_j ), g(a))$ can be made as small as we please by choosing $j$ sufficiently large. A similar remark holds for a sequence $(y_n)$ in $S$ which converges to $b$. From this show that $x_j$ is arbitrarily close to $y_k$ (for large $j$ and $k$) provided we assume that $a$ is sufficiently close to $b$. Use this in turn to show that $g(a)$ is arbitrarily close to $g(b)$ when $a$ and $b$ are sufficiently close.

The uniqueness argument is very easy.

Answer 2

(I also assume that we work with metric spaces.)
If you only need continuity of $g$, then continuity (not uniform) of $f$ is enough*. Let a sequence $x_n \in X$ converge to some point $x$. We will show that $g(x_n)$ converges to $g(x)$. I will write '(#)' in places, where we use the definition of $g$ for better readibility.

For every $x_n$ ($n$ is fixed) there exists a sequence $(w_n^m)_{m=1}^\infty \in D$ that converges to $x_n$. Since $f(w_n^m) \ \to g(x_n)$ (#) and $w_n^m \to x_n$, we may choose such $m_n$ that $d_X(w_n^{m_n}, x_n)$ and $d_Y(f(w_n^{m_n}), g(x_n))$ are smaller then $\frac{1}{n}$.
Now (from $d_X(w_n^{m_n}, x_n) < \frac{1}{n}$) you have: $\lim_{n \to \infty} w_n^{m_n} = x $, so (#):
$\lim_{n \to \infty} f(w_n^{m_n}) = g(x) $,
but we have also $d_Y(f(w_n^{m_n}),g(x_n)) < \frac{1}{n}$, so:
$\lim_{n \to \infty} g(x_n) = g(x) $, which gives us the continuity of $g$.

*However you still need uniform continuity to show that $g$ is uniquely defined, so my approach is not better, it is just different.