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Then can we say that Int(f(A)) a subset of f(Int(A))? Can we prove or provide counterexample for this claim?

The space is (R,d), where d is the absolute value metric, and R is the set of real numbers, f is a continuous function from R to R. A be a subset of R.

If yes, can you extend in for any arbitrary metric spaces I am very new to metric space and topology, so any help will be appreciated.

Doubts
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    You asked the same question with the closure today. The given answers to your previous question should have given you some insights on how approach such a question. So try to answer this question by yourself, it will be far more useful for you. – TheSilverDoe Oct 16 '20 at 13:58

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Yes, it is true. The following argument works for any continuous function $f\colon X\to Y$ between two topological spaces. If $f(x) = y\in \text{int}(f(A))$, then there is an open set $U$ such that $y\in U\subseteq f(A)$. By continuity, $f^{-1}(U)$ is open and $x\in f^{-1}(U)\subseteq A$, so $x\in \text{int}(A)$, i.e., $y\in f(\text{int}(A))$.

If you are not familiar with a general topological space, then you can take $X, Y$ to be metric spaces and $U$ to be an open ball around $y$. The inverse of an open ball is open by the usual $\epsilon-\delta$ definition.

red whisker
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