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Let X and Y be topological spaces, and let $f$ : $X$ → $Y$ be a function. I need to prove that the following definitions are equivalent:

(1) $f$ is continuous

(2) $∀A⊆X$, $f(Cl_X(A)) ⊆ Cl_Y(f(A))$

(3) $∀C⊆Y$ closed, $f^{-1}(C)$ is closed in $X$

(4) $∀V ⊆ Y$ open, there exists $U$ ⊆ $X$ open such that $f(U) ⊆ V$

I've proved the following equivalences: (1) = (3), (1) = (4). I want to show that (2) = (3) to complete the problem, but I don't quite get how to demonstrate the equivalence. I tried to use proof by contradiction but was unsuccessful, and I haven't been able to come up with a direct proof. Any advice?

(Note: $Cl_X(A)$ denotes the closure of $A$ in $X$).

Marcus Russell
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Assume (2) is true. Let $C \subseteq Y$ is closed. Then by (2), $f(Cl_X(f^{-1}(C))) \subseteq Cl_Y(f(f^{-1}(C))) \subseteq Cl_Y(C) = C$. Thus, $Cl_X(f^{-1}(C)) \subseteq f^{-1}(C)$, so $f^{-1}(C)$ is closed.

Conversely, assume (3) is true. Let $A \subseteq X$. Then $Cl_Y(f(A))$ is closed, so by (3), $f^{-1}(Cl_Y(f(A)))$ is closed. Clearly $A \subseteq f^{-1}(Cl_Y(f(A)))$, so $Cl_X(A) \subseteq f^{-1}(Cl_Y(f(A)))$. Hence, $f(Cl_X(A)) \subseteq Cl_Y(f(A))$.

David Gao
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  • By the way, as written, condition 4 is not equivalent to continuity. You probably want for any $x \in X$, there exists open $U \subseteq X$ which contains $x$ s.t. $f(U) \subseteq V$. – David Gao Dec 26 '23 at 05:45