Say that a metric space \(X\) has property (A) if the image of every continuous function \(f:X\to\mathbb R\) is an interval, which may be open, closed or half-open. Prove that \(X\) has property (A) if and only if it is connected.
Solution
1. If \(X\) is connected, then \(X\) has property (A).
Let \[ f:X\longrightarrow\RR \] be continuous. A continuous image of a connected space is connected, so \(f(X)\) is a connected subset of \(\RR\). The connected subsets of \(\RR\) are exactly the intervals. Hence \(f(X)\) is an interval, and \(X\) has property (A).
2. If \(X\) is disconnected, then \(X\) does not have property (A).
If \(X\) is disconnected, there are nonempty disjoint open sets \(U,V\subseteq X\) with \[ X=U\cup V. \] Since each is the complement of the other, both \(U\) and \(V\) are also closed. Define \[ f(x) \coloneqq \begin{cases} 0,&x\in U,\\ 1,&x\in V. \end{cases} \] The map \(f:X\to\RR\) is continuous because it is constant on the two clopen pieces \(U\) and \(V\). Its image is \[ f(X)=\{0,1\}, \] which is not an interval. Thus property (A) fails.
3. Therefore\[ \boxed{ X\text{ has property (A)} \iff X\text{ is connected}. } \]
Step 1 proves that connectedness implies property (A), while step 2 proves the contrapositive of the converse.
4. Q.E.D.
Step 3 is the required equivalence.