Let \(f: X \to Y\) be continuous. Show that if \(X\) is Lindelöf, or if \(X\) has a countable dense subset, then \(f(X)\) satisfies the same condition.
Solution
Write \(Z=f(X)\) with the subspace topology.
If \(X\) is Lindelöf and \(\{U_i\}_{i\in I}\) is an open cover of \(Z\), then \(\{f^{-1}(U_i)\}_{i\in I}\) is an open cover of \(X\). Choose a countable subcover \(f^{-1}(U_{i_1}),f^{-1}(U_{i_2}),\dots\). Surjectivity of \(f:X\to Z\) implies \(U_{i_1},U_{i_2},\dots\) cover \(Z\). Hence \(Z\) is Lindelöf.
If \(D\subset X\) is countable and dense, then \(f(D)\) is countable. It is dense in \(Z\): if \(W\subset Z\) is nonempty open, then \(f^{-1}(W)\) is nonempty open in \(X\), so it meets \(D\); hence \(W\) meets \(f(D)\). Thus \(Z\) is separable.