Let $f : X \to Y$ be a continuous map between metric spaces. Then $f(X)$ is a complete subset of $Y$ if
A. the space $X$ is compact
B. the space $Y$ is compact
C. the space $X$ is complete
D. the space $Y$ is complete.
My Attempt.
If I want to show completeness then I assume a cauchy sequence of $f(X)$ say $y_1,y_2,y_3,y_4 \dots \dots$ be a cauchy sequence in $f(X)$ let $y_1=f(x_1),y_2=f(x_2),y_3=f(x_3),y_4=f(x_4) \dots \dots$
Then how do we proceed futher systematically?
I have following ideas but I cannot use/implement properly. we want $y=f(x)$ where $(y_m) \to y$ and $(x_m) \to x$ but I wonder how to use them because I don't know $(x_m)$ converge or not.
A. is the correct answer since the continuous image $f(X)$ of a compact set $X$ is always compact, and hence complete.
B. C. D. are all false. Indeed, we can take $X=\Bbb R$ which is complete and take $Y=[-\pi/2,\pi/2]$ which is both compact amd complete. However, for $f(x)=\arctan(x)$ we have $f(X)=(-pi/2,\pi/2)$ which is not complete.