# show that $f$ is continuous

Show that $f:\mathbb{R} \rightarrow \mathbb{R}$ is continuous in the $\delta-\epsilon$ definition of continuity if and only if for all $x \in \mathbb{R}$ and all open set $U$ where $f(x) \in U$, there exists an open set $V$ where $x \in V$ and $f(V) \subset U$

Forward direction: Let open set $V$ such that $V \subset B_{\delta}(x)$. Then we have $f(V) \subset f(B_{\delta}(x)) \subset B_{\epsilon}(f(x)) \subset U$. The last inclusion follows by the openness of $U$ and also $f(x) \in U$

Backward direction: No idea. Can anyone guide me?

#### Solutions Collecting From Web of "show that $f$ is continuous"

Let $x$ be such that $f(x) \in U$. Then by assumption that $U$ is open, there is some $\epsilon > 0$ such that $B_{\epsilon} f(x) \subset U$. Find $\delta >0$ such that $y \in B_{\delta}(x) \implies f(y) \in B_{\epsilon}f(x)$ (why can we find such a $\delta$?). Now, take $V = B_{\delta}(x)$.