- #1
Fantini
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Hello. I wish to prove this:
$$\text{A function } f: X \to Y \text{ is continuous if and only if the inverse image of any closed set is closed.}$$
Proof: $(\implies)$ Let $V \subset Y$ be a closed se. By definition, $Y-V$ is an open set, and by the continuity of $f$ it follows that $f^{-1}(Y-V)$ is open in $X$. Thus, $f^{-1}(Y-V) = f^{-1}(Y) - f^{-1}(V) = X - f^{-1}(V)$ is open, and therefore $f^{-1}(V)$ is closed.
For the second implication, let $p \in X$ and $C$ a closed set in $Y$ such that $f(p) \notin C$. Thus $f(p) \in Y - C = V$, and by hypothesis $f^{-1}(C)$ is closed in $X$, with $p \notin f^{-1}(C)$. Hence $p \in X - f^{-1}(C) = U$, which is an open set. Therefore, we have an open set $U$ in $X$, with $p \in U$ and $f(p) \in V$ open in $Y$, such that $f(U) \subset V$. $\blacksquare$
My restlessness lies in the second implication. Is it right to assume that if $f(p) \notin C$ then $p \notin f^{-1}(C)$? Thanks. (Yes)
Fantini
$$\text{A function } f: X \to Y \text{ is continuous if and only if the inverse image of any closed set is closed.}$$
Proof: $(\implies)$ Let $V \subset Y$ be a closed se. By definition, $Y-V$ is an open set, and by the continuity of $f$ it follows that $f^{-1}(Y-V)$ is open in $X$. Thus, $f^{-1}(Y-V) = f^{-1}(Y) - f^{-1}(V) = X - f^{-1}(V)$ is open, and therefore $f^{-1}(V)$ is closed.
For the second implication, let $p \in X$ and $C$ a closed set in $Y$ such that $f(p) \notin C$. Thus $f(p) \in Y - C = V$, and by hypothesis $f^{-1}(C)$ is closed in $X$, with $p \notin f^{-1}(C)$. Hence $p \in X - f^{-1}(C) = U$, which is an open set. Therefore, we have an open set $U$ in $X$, with $p \in U$ and $f(p) \in V$ open in $Y$, such that $f(U) \subset V$. $\blacksquare$
My restlessness lies in the second implication. Is it right to assume that if $f(p) \notin C$ then $p \notin f^{-1}(C)$? Thanks. (Yes)
Fantini
