Let \(f: X\to Y \) be a continuous map between metric spaces. Then \(f(X)\) is a complete subset of \(Y\) if

A. X is compact

B. Y is compact

C. X is complete

D. Y is complete


Let \(X\) be compact. Then \(f(X)\) is compact. (continuous image of compact space is compact)

Now, compact subset of any Hausdorff space is closed. So in particular, compact subset of any metric space is closed.

Let \((y_n)\) be a Cauchy sequence in \(f(X)\). Then since \(f(X)\) is compact, \((y_n)\) has a convergent subsequence (converging to a point in that compact set i.e, in \(f(X)\) ).

Suppose \(y_{n_k} \to y \in f(X) \).

Then by triangle inequality, we have \(d(y_n,y) \le d(y_n,y_{n_k}) + d(y_{n_k},y) \to 0+0=0 \) as \(n\to \infty\)

Here we have used that \(y_n\) is cauchy to conclude \(d(y_n,y_{n_k}) \to 0 \).

So this implies \(y_n \to y\). Since \(y\in f(X)\) we conclude that \(f(X)\) is complete.

This proves A.

Let \(Y=[0,2]\) and \(X=(0,1)\). Take the inclusion map \(i(x)=x\) for all \(x\in X\). This example shows that even if we take \(Y\) to be compact, or complete, \(f(X)\) need not be complete. So this disproves B and D.

Now take \(X=\mathbb{R}\) and take \(Y=(0,1)\). We know there is a homeomorphism between these two sets where the metric is usual topology. So, in this case, the image of a complete set is not complete. This disproves option C.