Difference between continuous and uniformly continuous functions on a dense metric subspace.

Question

Let $X$ be a dense subset of metric space $(\tilde X,d)$. Let be $(Y,d')$ be a complete metric space and $ f: X \rightarrow Y$ a continuous mapping.

It follows from density that for all points in $\tilde X$ there is (at least one) sequence in $X$ that converges to it. Thus one can tentatively define an extension $\tilde f : \tilde X \rightarrow Y$ of $f$ by:

$\forall \tilde x \in \tilde X: \tilde f ( \tilde x) = \lim_{n \rightarrow ∞} f(x_n) $

where $(x_n)_{n \geq 1}$ is a sequence in $X$ converging to $\tilde x$.

Now, $\tilde f$ is easily seen to be well defined if $f$ is uniformly continuous. Is this still true if $f$ is just continuous? If not, please give or outline proof.

Answer 1

No.

$X=(0,1]$

$\overline{X}=[0,1]$

$Y=\mathbb{R}$

$f:X \rightarrow Y \quad \quad$ $\displaystyle x \mapsto \sin(\frac{1}{x})$