Nearest point property and uniformly continuous image

Question

Show that a uniformly continuous image of a metric space that has the nearest point property need not have that property.

I have some trouble understanding the problem. With the term ''uniformly continuous image'' does it mean that $f(X)=Y$ where $f$ is uniformly continuous ?

Answer 1

I did not know this nearest point property, but I found it in Google books and there the author gives the following equivalent formulations:

• Every infinite bounded subset of $X$ has an accumulation point in $X$.
• Every bounded sequence has a subsequence that converges in $X$.
• $X$ is complete and every bounded subset is totally bounded.

Now consider $X = \mathbb{N}$ in the usual metric, which has the nearest point property. And note that $f(n) = \frac{1}{n}$ is uniformly continuous onto a subspace of $\mathbb{R}$ that does not have that property.