I came across this problem in MIT course archives. I am currently trying to learn real analysis and the problem is as follows:
Let $(S, d_S )$ be a discrete metric such that $d_S(t, r) = 1;\ \forall t \ne r)$
(a) Show that any map $f : S\to X$ into another metric space $X$ is continuous; using the definition of continuity by sequences.
(b) Show that any map $f : S\to X$ into another metric space $X$ is continuous; using the definition of continuity by $\epsilon\delta$-balls.
(c) Which maps $f : \mathbb R\to S$ are continuous? (Give an easy characterization and prove it.)
My attempt:
A. Fix a point $x_\circ\in X$. Define a sequence $S=\{x_n\}$ such that $\lim\limits_{n\to\infty}x_n=x_\circ$ It follows $f(x_n)\in f(S);\forall \ n\in \mathbb{N}$. $S$ is open.
If I can prove $f(S)$ is open, it would imply $f$ is continuous. If $f(S)$ doesn't contain{how do you write this in MathJax} $f(x_\circ)$, it is open. Correspondingly, $S$ is open. Hence, $f:S\to X$ is continuous.
B. Here is my argument using $\epsilon-\delta$ definition. Fix a point $x_\circ$. $d_S(p,x_\circ)=1<\delta\ \ \forall \ \ p\ne x_\circ\in S$. We can pick $\epsilon=\sup\{d_X(f(x),f(x_\circ)\}$
C. My guess is only constant functions will be continuous.
Please check.
Best Answer
A. Proving that $f(S)$ is open prove nothing. If $(x_n)_{n\in\mathbb N}$ converges to $x_0$ in $S$ then there is some $N\in\mathbb N$ such that $n\geqslant N\implies d(x_n,x_0)<1$. Therefore,$$n\geqslant N\implies x_n=x_0\implies f(x_n)=f(x_0)\iff d\bigl(f(x_n),f(x_0)\bigr)=0.$$So, $\lim_{n\to\infty}f(x_n)=f(x_0)$.
B. Let $\varepsilon>0$ and take $\delta=1$. Then$$d(x,x_0)<\delta\iff x=x_0\implies d\bigl(f(x),f(x_0)\bigr)=0<\varepsilon.$$
C. Yes, only constant functions will be continuous. That's because $\mathbb R$ is connected and therefore $f(\mathbb R)$ is connected too. But the only non-empty connected subsets of a discrete metric space are the singletons.