For any two metric spaces$$(X, d_X), (Y, d_Y)$$​, a map f: X → Y is said to be a closed map if whenever F is closed in X,

then f(F) is closed in Y. For any subset B of a metric space, B is given the induced metric.

The metric on X × Y is given by

​$$d((x,y), (x',y')) = max\{d_X(x, x'), d_Y(y, y')\}$$​​

Which of the following are true?

Correct option is C

Option A : Let A= (0,1) then  inclusion map $$i:A\to \mathbb{R}$$ will be open.

Hence, Option A is incorrect.

Option B:

Consider $X=Y=\mathbb{R}$, and the closed set $F$ in $X\times Y$ given by:​​

F={(x,y)∈R2∣xy=1}F = \{ (x, y) \in \mathbb{R}^2 \mid xy = 1 \}F={(x,y)∈R2\mathbb{R}^2R2​∣xy=1}

The projection onto $X$ is:

​$$p_1(F)=\{ x:x\neq0\}$$​​

which is not closed in $\mathbb{R}$ since it is missing $x=0$.

Option C : If $g\circ f$ is Closed, then g∣f(X)g |_{f(X)}g∣f(X)​ is Closed

• Given that $g\circ f:X\to Z$ is closed, does it imply that the restriction g∣f(X)g |_{f(X)}g∣f(X)​ is closed?
• Yes! If $g\circ f$ is closed, then for every closed set $F\subseteq X$, $g(f(F))$ is closed in $Z$,
• meaning that $g$ behaves as a closed map when restricted to $f(X)$.

Hence, Option C is correct.

Option D :

Consider $X=\mathbb{R}$, $Y=\mathbb{R}$, and the function:

​$$f(x) =\begin{cases} 1, & x > 0 \\[10pt] 0, & x \leq 0\end{cases}$$​​

Here, closed balls in $X$ map to closed sets in $Y$.

But $f$ is not closed because the closed set $F=\{0\}$ in $X$ maps to $\{0\}$, while $F=(-\infty ,0]$ maps to $\{0\}$,

which is not closed under $f$.