# Thread: Inverse of an open set is open implies continuous

1. ## Inverse of an open set is open implies continuous

A theorem in my book says:

Suppose that $(X_1 , \rho \ , \ (X_2, \gamma )$ are metric spaces. Define a function $f: X_1 \rightarrow X_2$. Then f is continuous iff $f^{-1} (U)$ is open in $X_1$ for every open set $U \subset X_2$.

Proof in the book goes:

Assume that for each open $U \subset X_2$, $f^{-1}(U)$ is open in $X_1$.

Let $x \in X_1$, then $f(x) \in X_2$.

Then $\forall \epsilon > 0$, $f^{-1} (B(f(x), \epsilon ))$ is open in $X_1$ since $B(f(x), \epsilon )$ is open in $X_2$, therefore f is continuous.

Q1: I understand why the set in the inverse is open, but how does that implies f is continuous? In other words, if I fill in the details, say:

Let $x_0 \in X_1$, then $\forall \epsilon > 0$, pick $\delta > 0$, then whenever $\rho (x_0, x ) < \delta \ \ \ \ \ for \ x \in X_1$, I need to get $\gamma ((f(x_0),f(x))) < \epsilon$ But how did the statement from above help me to get there?

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Conversely, suppose that f is continuous, let U be open set in $X_2$.

$\forall y \in U$, there exists $\epsilon _y > 0 \ s.t. \ B( \epsilon _y , y ) \subset U$,
and $\forall x \in f^{-1}( \{ y \} ) , \exists \delta _x > 0$

such that $B( \delta _x , x ) \subset f^{-1}(B( \epsilon _y , y ) ) \subset f^{-1} (U)$

Implies that $f^{-1}(U) = \bigcup _{x \in f^{-1}(U) } B( \delta _x , x )$ is open.

Q2: I understand that the open ball around y is in U since it is an open set, why I'm a bit unsure about how to get $B( \delta _x , x) \subset f^{-1}(B( \epsilon _y , y ))$, I'm sure it is because f is continuous. I know that continuous function map an open set to an open set, but do I know for sure it is true for the inverse?

Thank you!

A theorem in my book says:

Suppose that $(X_1 , \rho \ , \ (X_2, \gamma )$ are metric spaces. Define a function $f: X_1 \rightarrow X_2$. Then f is continuous iff $f^{-1} (U)$ is open in $X_1$ for every open set $U \subset X_2$.

Proof in the book goes:

Assume that for each open $U \subset X_2$, $f^{-1}(U)$ is open in $X_1$.

Let $x \in X_1$, then $f(x) \in X_2$.

Then $\forall \epsilon > 0$, $f^{-1} (B(f(x), \epsilon ))$ is open in $X_1$ since $B(f(x), \epsilon )$ is open in $X_2$, therefore f is continuous.

Q1: I understand why the set in the inverse is open, but how does that implies f is continuous? In other words, if I fill in the details, say:

Let $x_0 \in X_1$, then $\forall \epsilon > 0$, pick $\delta > 0$, then whenever $\rho (x_0, x ) < \delta \ \ \ \ \ for \ x \in X_1$, I need to get $\gamma ((f(x_0),f(x))) < \epsilon$ But how did the statement from above help me to get there?

---------------

Conversely, suppose that f is continuous, let U be open set in $X_2$.

$\forall y \in U$, there exists $\epsilon _y > 0 \ s.t. \ B( \epsilon _y , y ) \subset U$,
and $\forall x \in f^{-1}( \{ y \} ) , \exists \delta _x > 0$

such that $B( \delta _x , x ) \subset f^{-1}(B( \epsilon _y , y ) ) \subset f^{-1} (U)$

Implies that $f^{-1}(U) = \bigcup _{x \in f^{-1}(U) } B( \delta _x , x )$ is open.

Q2: I understand that the open ball around y is in U since it is an open set, why I'm a bit unsure about how to get $B( \delta _x , x) \subset f^{-1}(B( \epsilon _y , y ))$, I'm sure it is because f is continuous. I know that continuous function map an open set to an open set, but do I know for sure it is true for the inverse?
Then you know something that is not true! For example, f(x)= a, that is, the function that is maps all x to the fixed point a, is certainly continuous but f(U) for any set U, including open sets, is the singleton set {a} which is NOT open. I suspect you are simply remembering, incorrectly, the fact that the continuous inverse image of an open set is open.

Thank you!
Every ball, $B(\epsilon_y,y)$ is open, of course, so $f^{-1}(B(\epsilon_y, y))$ is open. That means that every point in the set, in particular x itself, is an "interior point" and so there exist neighborhood $B(\delta, x)\subset f^{-1}(B(\epsilon_y, y)$. That is all you need.