I’m not a mathematician and I’m very new to topology, I’m just a guy who’s curious about shapes.
My favorite function is the Weierstrass function, which is continuous everywhere but differentiable nowhere. Is its curve connected? Is that even a valid question to ask for a non-differentiable curve?
Thanks for your time.
Yes. More generally, the image of any continuous function from an interval to $\mathbb{R}$ will be path connected: let $f : I \subseteq \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function on an interval $I$ (such as the whole real line) and let $\mathcal{C} := im(f)$ be its image. Now, pick any two points $y,y' \in \mathcal{C}$ so that there exist $x,x' \in I$ with $f(x) = y$ and $f(x') = y$. Since $I$ is an interval and $x,x' \in I$, we have that $[x,x'] \subseteq I$ and thus we can take the following continuous path:
$$ \gamma : [x,x'] \to \mathcal{C} \\ t \to f(t) $$
which verifies $\gamma(x) = f(x) = y$ and $\gamma(x') = f(x') = y'$ as desired.
Moreover, any continuous function from a (path) connected space will have a (path) connected image.
For path connectedness, the proof is the following: let $f: C \to D$ be a continuous function with $C$ path connected and let $p,q \in im(f)$ with $p = f(a)$ and $q = f(b)$. By path connectedness of $C$, we have a path $\gamma$ from $a$ to $b$. Thus, $f \circ \gamma$ is a path from $p$ to $q$.
The connected case requires a bit more topology (but not that much).
Yes, the graph of the function is connected.
Let $f \colon [0,1] \to \mathbf{R}$ be any continuous function.
Then the mapping $F \colon [0,1] \mathbf{R}^2$ defined by $F(t) = (t,f(t))$ is itself continuous. The graph in question is the image of $[0,1]$ under the mapping $F$. But the image of a connected set under a continuous mapping is connected. Since $[0,1]$ is connected, the graph is connected.
