Every function from discrete metric space to another metric space is uniformly continuous

Question

My solution:It is fairly straightforward graphically but I just want to ensure if it is rigorous enough.

Suppose $X$ is a discrete metric space and $f$ be any function from $X$ to $Y$ where $Y$ is any other metric space. Let $d_{x}$ and $d_{y}$ denote the metrics in metric spaces $X$ and $Y$ respectively.

If $p,q\in X, p=q$, then $d_{x}(p,q)=0<\delta$ for any $\delta>0$. If $p,q\in X, p\neq q$, then $d_{x}(p,q)=1$.

Choosing $\delta=2$, we have $d_{y}(f(p),f(q))<\epsilon$, $\forall \epsilon>0$.

So $\forall \epsilon>0,$ we have $\delta>2$, such that $d_{x}(p,q)<\delta$ implies $d_{y}(f(p),f(q))<\epsilon$.

So, any function from a discrete metric space to any other metric space is uniformly continuous.

Please suggest if it is correct and rigorous enough.

Answer 1

Hints:

For any $\epsilon>0$, you can choose that $\delta=\frac12$, then for any $x,y \in X$, if $|x-y|<\delta$, then $x=y$, and hence $|f(x)-f(y)|=0<\epsilon$, which shows that $f(x)$ is uniformly continuous.

Answer 2

Choosing $\delta=2$, we have $d_{y}(f(p),f(q))<\epsilon$, $\forall \epsilon>0$.

No, that is wrong, if you choose $\delta = 2$ then $p$ and $q$ can be ANY points in $X$, because the distance between any two points in $X$ is always less than 2. So that won't tell you anything about $d(f(p), f(q))$ at all. If you set $\delta<1/2$ then you force $p=q$ thus obviously $f(p)=f(q)$ so you have uniform continuity.

Answer 3

To show that $f$ is continuous:

Discrete metric induces discrete topology, then preimage of any open set in $Y$ is automatically open in $X$ thus $f$ is continuous. (For uniform continuity use the fact that singletons are open as Paul and Suzu. Hirose pointed out earlier.)