Let $\mathbf{Top}$ be the category of topological spaces, and let $I\in\mathbf{Top}$ be the unit interval $I=[0,1]\subset\mathbb{R}$. For any space $X$, let $|X|$ denote the underlying set of points; similarly if $f\colon X\to Y$ is continuous, $|f|$ denotes the underlying function. Define a *curve in $X$* to be a continuous map $I\to X$.

Given spaces $X$ and $Y$ and a function $f\colon |X|\to |Y|$ on underlying sets, say that $f$ *sends curves to curves* if composing $f$ with any curve $c\colon I\to X$ yields a (continuous) curve in $Y$. Of course, if $f$ is continuous then it sends curves in $X$ to curves in $Y$, but the converse may not hold.

I'll say that $X$ *has enough curves* if, for any $Y$ and function $f\colon|X|\to|Y|$, we have that $f$ is continuous if and only if it sends curves to curves.

An example of a space that does not have enough curves is the "sequence space" $S=\{0\}\cup\{\frac{1}{n+1}\mid n\in\mathbb{N}\}\subset I$. All curves in $S$ are constant, so they cannot detect non-continuous maps out of $S$.

**Question:** Which well-known classes of spaces have enough curves?

The Convenient Setting of Global Analysis(AMS, 1997), $X$ has the final topology with respect to continuous curves if $X$ is the strong dual of a Fréchet-Montel locally convex vector space. Such spaces are far from being discrete - take e.g. $X=\mathbb{R}^n$ or $X=$ the space of distributions of compact support on a second countable smooth manifold. $\endgroup$ – Pedro Lauridsen Ribeiro Jun 16 '16 at 5:51