A <topological space> is <connected> if it cannot be expressed as a union of two disjoint nonempty sets open in that space. Suppose $f(X)$ had such a separation $U\cup V$. Here $U,V$ are open in the <subspace topology> of $f(X)$. Continuity makes $f^{-1}(U)$ and $f^{-1}(V)$ open in $X$. They are disjoint, nonempty because $U,V$ contain image points, and cover $X$. This contradicts connectedness. Hence \b[a <continuous image of a connected space> is connected].

Solution

Past exam of the metric spaces and complex analysis course of the University of Oxford / 2017 / 2 / a / i / Solution

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