Is there a way to phrase "continuous" in terms of "connected"?

Solution 1:

The short answer is that there are functions which are not conventionally continuous, but which are connected-set-contininuous.

For example, one can use any $f:\mathbb{R}\rightarrow \mathbb{R}$ with the following property:

If $X:=(a,b)$ is any non-empty open interval, then $f(X) =\mathbb{R}$.

One example of such an $f$ is Conway's base 13 function.

Such a function cannot be continuous because it is unbounded on every non-trivial closed interval (because a non-trivial closed interval contains a non-trivial open interval as a subset).

On the other hand, the connected subsets of $\mathbb{R}$ are points or are intervals (with or without some end points). Of course, the image of a point is connected. The image of an interval is also connected: every interval has an open subinterval, and then $f$ applied to the open subinterval is already all of $\mathbb{R}$, which is connected.

Solution 2:

For a counterexample, consider

$$ \begin{array}{l} f : \mathbb R \to \mathbb R \\ f(x) = \begin{cases} \sin(1/x) & \mbox{if } x\neq 0 \\ 0 & \mbox{otherwise} \end{cases} \end{array} $$ which is not continuous at $0$.

In spite of that, we can show that $f$ is connected-set-continuous, that is: if $X \subseteq \mathbb R$ is a connected set (i.e. an interval), then the image $f(X)$ is connected (an interval).

Indeed, if $0\not\in X$, then $f$ is continuous on $X$ and by the intermediate value theorem $f(X)$ is an interval.

If $X=\{0\}$ then $f(X)=\{0\}$ which is connected (a degenerate interval).

Otherwise, $X$ contains either $[0,\epsilon)$ or $(-\epsilon,0]$ for some $\epsilon>0$, hence $f(X)=[-1,+1]$.