 Solution

ID: past-exam-of-the-metric-spaces-and-complex-analysis-course-of-the-university-of-oxford/2014/1/b/iv/solution

True. For any continuous map,

f (\overline{A}) \subseteq \overline{f (A)}

f (x)

pulls back to an open neighbourhood of

x \in \overline{A}

, and therefore meets

A

in its preimage.

Apply the same inclusion to the continuous inverse of the homeomorphism:

f^{- 1} (\overline{f (A)}) \subseteq \overline{f^{- 1} (f (A))} = \overline{A} .

Applying

f

gives the reverse inclusion, proving

\overline{f (A)} = f (\overline{A}) .

The closures are taken in their respective ambient spaces.

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