Suppose that (a) is true and let $\left\{p_n\right\}\subset E$ be a sequence such that (c) holds. Let Then, for some if then Now, there is also some $N$ such that whenever $n\ge N,$ and thus, whenever $n\ge N,$ so (b) holds.

Conversely, suppose that (a) is false. Then, for some that for every there exists a point $x\in E$ such that $d_Y(f(x),q)\ge ϵ$ but If we let ${\delta }_n=1/n,n=1,2,3,\dots ,$ and each $x_n$ a point such that then $\left\{x_n\right\}$ is a sequence satisfying (c). However, since $d_Y(f(x_n),q)\ge ϵ$ for all $n,$ (b) does not hold.

This follows directly from the that limits of functions are characterized by limits of sequences, and the algebraic properties of sequences of limits.

Note that the definition of a function having a limit at point in a metric space is different from the definition of a function being continuous at a point in a metric space only in that the continuous definition requires the function to be defined at the point (and equal to the limit at the point.)

Let Since $g$ is continuous at $f(p),$ there exists such that

Since $f$ is continuous at $p,$ there exists such that

It follows that

if and $x\in E.$ Thus, $h$ is continuous at $p.$

Assume $f$ is continuous on $X$ and $V$ is an open set in $Y.$ Suppose, for the sake of contradiction, that $f^{-1}(V)$ is not open. Then, some point $p\in f^{-1}(V)$ is not an interior point of $f^{-1}(V),$ which means there is no neighborhood of $p$ that contains only points in $f^{-1}(V),$ that is, every neighborhood of $p$ contains some point $q$ that is not in $f^{-1}(V),$ i.e., $f(q)\notin V.$ Now, since $V$ is open and $f(p)\in V,$ there is some for which $B_{ϵ}(f(p))\subset V,$ but, since all neighborhoods of $p$ contain some $q\notin f^{-1}(V),$ there is no for which all points within $\delta$ of $p$ are mapped to $B_{ϵ}(f(p))$ by $f,$ a contradiction, since $f$ is continuous on $X$ by hypothesis. Therefore, our assumption is incorrect and $f^{-1}(V)$ is open.

Conversely, suppose $f^{-1}(V)$ is open in $X$ for every open set $V$ in $Y.$ Let $V$ be an open set in $Y.$ Assume, for the sake of contradiction that there is some $p\in f^{-1}(V)$ at which $f$ is not continuous. Let Then, there is no for which $B_{\delta }(p)$ contains only points that are mapped by $f$ to $V,$ i.e. every neighborhood of $p$ contains some point that is not in $f^{-1}(V),$ and therefore $p$ is not an interior point of $f^{-1}(V),$ and $f^{-1}(V)$ is not open, a contradiction. Thus, our assumption must be incorrect, and there is no such $p,$ and since entire space $Y$ is an open subset of itself, $f$ must be continuous on all of $X.$