= Solution
For any topological target $Y$, \b[$f(K)$ is compact]. Pull an open cover of $f(K)$ back by $f$, take a finite subcover of $K$, and map the covering statement forward. Without a separation assumption on $Y$, the image need not be closed, as a constant map into a nonclosed singleton of an indiscrete space shows.
If $Y$ is a <metric space>, $f(K)$ is also closed in $Y$, bounded, <totally bounded>, complete with its inherited metric, and sequentially compact. Closedness uses the Hausdorff theorem from part (iii); total boundedness follows by taking finite subcovers of the covers by balls of any fixed radius. Compact metric spaces are sequentially compact, and a <Cauchy sequence> with a convergent subsequence converges to the same limit, proving <completeness>. These properties are asserted for the image subset, without requiring the whole ambient <metric space> to be complete.
Back to article page
