Given topological space $\mathcal{X}$ and $\mathcal{Y}$, a function $f:\mathcal{X}\rightarrow\mathcal{Y}$ is continuous if the inverse image $f^{-1}(V)=U$ of an open set $V$ of $\mathcal{Y}$ is an open set $U$ of $\mathcal{X}$. Note how this generalizes the notion of a continuous function from real analysis - $f:\mathbb{R}\rightarrow \mathbb{R}$ is continuous at $a$ if $f(a)=\lim_{x\rightarrow a}f(x)$. In other words, a function is continuous when points near $a$ give output near $f(a)$. Table source $$\begin{array}{|c|c|} \hline \text{Continuity on } \Bbb R & \text{Topological space} \\ \hline \text{Fix } \varepsilon & \text{Fix a neighborhood } V \text{of } f(x_0) \\ \hline \text{exists } \delta & \text{exists a neighborhood } U \text{of } x_0\\ \hline |x-x_0|\le \delta \Rightarrow |f(x)-f(x_0)|\le \varepsilon & f(U)\subseteq V \\ \hline \end{array}$$