Proof. Suppose, on the contrary, that \(f(E) = A\cup B\), where \(A\) and \(B\) are non-empty separated sets. Put \(G = E \cap f^{-1}(A)\) and \(H = E\cap f^{-1}(B)\). Then \(E = G \cup H\), and neither \(G\) nor \(H\) is empty.

We show that \(\overline {G}\cap H = \emptyset \). Let \(x\in \overline {G}\) with \(x\in H\). Then \(x\in E\subseteq X\), and \(x\) is the limit of a sequence \(\{g_n\}\) in \(G\). By continuity \(f(g_n)\longrightarrow f(x)\), and \(f(g_n)\in A\), so \(f(x)\in \overline {A}\). But \(f(x)\in B\), and \(\overline {A}\cap B = \emptyset \), a contradiction. The same argument shows that \(G\cap \overline {H} = \emptyset \).

Thus \(G\) and \(H\) are non-empty separated sets with union \(E\), contradicting the connectedness of \(E\). Therefore \(f(E)\) is connected. □

Solution. The interval \([0,1]\) and the line \(\mathbb {R}\) are connected, so their continuous images are connected, that is, intervals. The set \(\{0,1\}\) is not an interval, and neither is \(\mathbb {Q}\) (it omits \(\sqrt {2}\), which lies between \(1\) and \(2\)). So no such functions exist. In particular, a continuous function on an interval that takes only rational values must be constant.

Solution. Let \(p(x) = a_nx^n + \dots + a_0\) with \(n\) odd and, without loss of generality, \(a_n>0\). For \(x\neq 0\), \[p(x) = a_nx^n\left (1 + \frac {a_{n-1}}{a_nx} + \dots + \frac {a_0}{a_nx^n}\right ),\] and the bracket tends to \(1\) as \(|x|\longrightarrow \infty \). Since \(n\) is odd, \(p(x)\longrightarrow +\infty \) as \(x\longrightarrow \infty \) and \(p(x)\longrightarrow -\infty \) as \(x\longrightarrow -\infty \). So \(p\) takes a negative value and a positive value. The image \(p(\mathbb {R})\) is connected, hence an interval, and so it contains \(0\). For example, \(x^3+x-1\) has a root between \(0\) and \(1\), and since it is strictly increasing the root is unique (about \(0.682\)).