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Continuity in terms of closed sets

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
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
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
Hello. I wish to prove this:Is it right to assume that if $f(p) \notin C$ then $p \notin f^{-1}(C)$?
Yes, it is right. By definition $p\in f^{-1}(C)\Leftrightarrow f(p)\in C.$