 Solution

ID: past-exam-of-the-probability-statistics-and-options-i-course-of-the-university-of-oxford/2009/e/2/a/ii/solution

Let

{U_{i}}

be an open cover of

f (X)

. By continuity the inverse images

f^{- 1} (U_{i})

are open and cover

X

. Compactness provides finitely many of them covering

X

, and the corresponding

U_{i}

then cover

f (X)

. Thus a continuous image of a compact space is compact.

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