Is $id$ an uniformly continuous function?
Let $(\mathbb{N},d)$ be a metric space such that $d(x,y) = |\frac{1}{x}-\frac{1}{y}|$.
Let us consider the identity map $id:(\mathbb{N},d) \to (\mathbb{N},d_{disc})$ such that $id(x)=x$. Then we see that this map is continuous.
We know that the sequence $\frac{1}{n}$ is a cauchy sequence under the usual metric. We choose $\epsilon =1$ then we see that there exists $\mathbb{N_1} \in \mathbb{N}$ such that $\frac{1}{N_1}-1 < \frac{1}{n} <\frac{1}{N_1}+1$ for all $n \ge N_1$.
Let us consider the set $B_d(N_1,1) = \{n \in \mathbb{N} : n \ge N_1\}$ . Then this set is closed as $(N,d)$ has a discrete topology. Then $B_d(N_1,1)$ is a closed and bounded subset of $\mathbb{N}$.
Then let us consider the function $id_{B_d(N_1,1)}$.If the function is uniformly continuous then we know that it maps cauchy sequences to cauchy sequences. Now consider the seqeunce $(n)$ where $n \ge N_1+1$ then $(n) \subset B_d(N_1,1)$ and it is also a cauchy sequence.
However, the sequence is not cauchy in $(N,d_{disc})$ as the only cauchy sequences in $(N,d_{disc})$ are the ones that are constant. Hence we know that $id_{B_d(\mathbb{N_1},1)}$ is not uniformly continuous.
Is my proof okay?
If this is okay can we consider $id$ as an example of a function which is continuous on a closed and bounded interval but not uniformly continuous. As the intervals of $\mathbb{N}$ under $d$ or under the usual induced metric is of the above form.