Increasing real valued function whose image set is connected

Let $S = [0,1) \cup [2,3]$ and $f\colon S \rightarrow \mathbb R$ be such that $f(S)$ is connected . Which of the following are true:

a) $f$ is discontinuous exactly at one point.

b) $f$ is discontinuous exactly at two points.

c) $f$ is discontinuous at infinite many points.

d) $f$ is continous.

I think $f$ is continuous . We need to show that $\lim_{x \to a^+}f(x) = f(a)$ for any $a \in S$. Since $f(x)$ is increasing, $\lim_{x \to a^+}f(x) \leq f(a)$. Suppose $\lim_{x \to a^+}f(x) < f(a)$. Then there is $l$ such that $\lim_{x \to a^+}f(x) < l <f(a)$, which contradicts connectedness of $f(S)$. Similarly we can show that $\lim_{x \to a^-}f(x) = f(a)$.

I would be thankful if anyone would check my solution and correct me if there are any errors.

Solution 1:

Your proof is pretty good but it has a couple of loose ends you should tie up. First, you assume $\lim_{x\rightarrow a^{+}}f(x)$ exists. It does, but you need to show it. One way would be to let $x_n$ be any sequence such that $x_n\to a$ and $x_n$ is monotone decreasing. Then $\langle f(x_n)\rangle$ is monotonically decreasing and is bounded below by $f(0)$. Therefore $\langle f(x_n)\rangle$ converges by the monotone convergence theorem. Thus you can conclude the continuous limit converges.

Next you should show why $\lim_{x\rightarrow a^{+}}f(x)<\ell<f(a)$ implies $f(S)$ is not connected. You can say for example that if $A=f(S)\cap(-\infty,\ell]$ and $B=f(S)\cap[\ell,\infty)$ then $S=A\cup B$ a disjoint union of closed sets in $f(S)$ which is impossible if $f(S)$ is connected.