Metric Space and Uniformly Continuous Functions Find an unbounded set $A$ such that every function from $A$ to a metric space is uniformly
continuous.
I'm drawing a blank for this problem. I've been just looking over the definitions on wiki, and am not really sure how to proceed here.
 A: We will begin by analysing some particular types of functions that must be (uniformly) continuous to help determine properties of such a set $A \subseteq \mathbb{R}$.


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*First, note that for each $x_0 \in A$ the function $f_{x_0} : A \to \mathbb{R}$ defined by $$f_{x_0} ( x ) = \begin{cases}1,&\text{if }x = x_0\\0,&\text{if }x \neq x_0\end{cases}$$ must be continuous (forgetting the uniformly part for the moment). Then $f_{x_0}^{-1} [ (\frac{1}{2} , \frac{3}{2} ) ] = \{ x_0 \}$ must be an open subset of $A$, which tells us that $x_0$ is an isolated point of $A$.
So every point of $A$ must be isolated, meaning that $A$ is a discrete subspace of $\mathbb{R}$.

*Now suppose that $A$ has an accumulation point $y$ (which by the above cannot belong to $A$). Then there must be a one-to-one sequence $\langle x_n \rangle_{n=0}^\infty$ in $A$ which converges to $y$.  Note that the function $f : A \to \mathbb{R}$ defined by $$f(x) = \begin{cases}
0, &\text{if }x = x_{2n}\text{ for some }n\\
1, &\text{otherwise}
\end{cases}$$ is uniformly continuous. So there is an $\delta > 0$ such that $| f(x) - f(x^\prime) | < \frac{1}{2}$ whenever $|x-x^\prime| < \delta$. Since the sequence $\langle x_n \rangle_{n=0}^\infty$ is Cauchy, there is an $N$ such that $| x_n - x_m | < \delta$ for all $n,m > 0$, and in particular $| x_N - x_{N+1} | < \delta$. But note that $| f(x_N) - f(x_{N+1}) | = 1$
Therefore $A$ has no accumulation points, and in particular $A$ is closed.
It shouldn't be too difficult to show that a familiar unbounded, closed, discrete subset of $\mathbb{R}$ (e.g., $\mathbb{Z}$) actually works.
