Assume that $f:\mathbb{R}^{2}\rightarrow \mathbb{R}^{2}$ is a continuous on a set $A$. Let $B \subset A$ be a generic set in $\mathbb{R}^{2}$ i.e, the countable intersection of the open and dense subset.



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> > Question: Suppose that the set of $f(x)$ for $x\in B$ is closed and convex set. Is the set of $f(x)$ for $x\in A$ closed and convex?
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My attempt: Since $B$ is a generic set, for every $x\in A$ there is a sequence $x\_{n} \in B$ convergents to $x$. Since $f$ is continuous, $f(x\_{n}) \rightarrow f(x)$. Then, I don't know how to get result.


If the answer is negative, then under which assumption it holds.



Closed on generic set implies closed set whole set
Obviously $f(B) \subseteq f(A)$. And since $B$ is dense in $A$, we have $A \subseteq \overline{B}$, and so $f(A) \subseteq f(\overline{B})$. But by the continuity of $f$, we have $f(\overline{B}) \subseteq \overline{f(B)}$. Since $f(B)$ is closed by assumption, this shows that $f(A) \subseteq f(B)$. So in fact we have $f(B) = f(A)$.


