Discussion
Let \(X\) and \(Y\) be metric spaces, and let \(f : X \to Y\) be a continuous map. For any subset \(S\) of \(X\), which one of the following statements is true?
Answer: B
The true statement is: if \(S\) is connected, then \(f(S)\) is connected.
A continuous image of a connected set is connected: if \(f(S)\) split into two disjoint relatively open pieces, their preimages would split \(S\) the same way.
The others fail. Open: \(f(x) = 0\) on \(\mathbb{R}\) sends the open set \(\mathbb{R}\) to \(\{0\}\), which is not open. Closed: \(f(x) = e^{x}\) sends the closed set \(\mathbb{R}\) to \((0, \infty)\), which is not closed. Bounded: \(f(x) = 1/x\) on \(X = (0, 1)\) sends the bounded set \((0, 1)\) to \((1, \infty)\). Continuity preserves compactness and connectedness, not openness, closedness or boundedness.
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