= Boundedness extends across closure under a continuous map
{title2=$f(A)\text{ bounded}\Longrightarrow f(\overline A)\text{ bounded}$}
For a <continuous map> into a <metric space>, $f(\overline A)\subseteq\overline{f(A)}$: an <open neighborhood> of an image point pulls back to a <neighborhood> meeting $A$. A <bounded set> lies in a <closed ball>, and its <closure> lies in the same ball because <distance> from the centre is a <continuous function>. The displayed boundedness conclusion follows. Neither <compactness> of the domain nor uniform <continuity> of the map is needed.
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