Continuous functions on metric spaces I am considering a problem (from Goldberg, "Methods of Real Analysis") where f is a continuous injective map from an arbitrary metric space, (M,d), to the discrete metric space (the reals with the discrete metric). The problem asks to show that for any pt a in M, {a} is an open ball in M.
I know how to prove this statement (f continuous implies the inverse image of any open ball about f(a) with diameter < 1 will contain an open ball about a, but this set, since f is injective, will be {a}, so {a} must be an open ball in M).
The problem is that it doesn't seem to make sense to me. If we choose (M,d) to be the reals with the absolute value function, then I can't see how {a} can be an open ball.
Thanks,
Matt
 A: Your proof is correct: for $a \in M$ there exists a $\delta > 0$ at $a$ for $\epsilon = 1$, so for all $x \in M$: $d(a, x) < \delta$ implies $d(f(x), f(a)) < 1$, and the latter means that $f(a) = f(x)$ and thus $x = a$. So $d(a, x) < \delta $ implies $a = x$, so $B(a, \delta) = \{a\}$.
So in fact if such a 1-1 continuous map exists, we have shown $(M,d)$ to have the discrete topology (not necessarily that $d$ is the discrete metric, of course), so your remark about $\mathbb{R}$ cannot apply: no such $f$ can exist. 
A: You have "a continuous injective map from an arbitrary metric space, $(M,d)$, to the discrete metric space (the reals with the discrete metric)".
That implies a few things!
One of the things it implies is that $(M,d)$ is not the reals with the usual metric, unless the continuous function is constant (but then it's not injective!).
What you need to look at what the hypotheses imply about the function $f$ and its domain, thinking of the domain not simply as a set but as a metric space.  If $f$ is not constant, then what do the hypotheses imply about the nature of the space $(M,d)$?
