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Is there $f:\mathbb{R}\to\mathbb{R}$ such that the graph is connected but for any two points in the graph, there isn´t any path in the graph between them?

I´m not sure what the answer is going to be. While trying to prove that there is such a function, it only occurred to me using functions such that the image of any open set is all the real line, but that seems not to be enough to ensure that the graph is connected.

Edit: The function exists, as shown in the comments of Brandon du Preez and HallaSurvivor. In fact, there are solutions $f$ with $f(x)+f(y)=f(x+y)\forall x,y$.

Saúl RM
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    This may be of interest: https://math.stackexchange.com/questions/3273874/can-a-nowhere-continuous-function-have-a-connected-graph There's a link to a paper in the accepted answer which constructs nowhere continuous functions with connected graphs. Haven't checked to see if they are path-connected or not though. – Brandon du Preez Jan 16 '22 at 09:05
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    In this answer, Eric Wofsey shows (for functions $f : \mathbb{R} \to \mathbb{R}$) that the graph of $f$ over $X \subseteq \mathbb{R}$ is path connected for each path connected $X$ if and only if $f$ is continuous. Combining this with Brandon's comment regarding nowhere continuous functions whose graphs are connected should give the claim – HallaSurvivor Jan 16 '22 at 09:17
  • Instead of editing your question to indicate a solution, you should answer your own question! That way other users on the site know how to spend their time ^_^ – HallaSurvivor Jan 16 '22 at 18:28

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As explained in the comments by Brandon du Preez and HallaSurvivor, such functions exist. Indeed, the following paper:

F. B. Jones, Totally discontinuous linear functions whose graphs are connected, November 23, (1940).

proves the existence of a function $f:\mathbb{R}\to\mathbb{R}$ which has a connected graph but it is totally discontinuous.

If you pick distinct points in the graphs of $f$, $(x,f(x))$ and $(y,f(y))$, with $x>y$, there can be no path between them in the graph. This is because if there was a path between them, there would be an arc $c:[0,1]\to\mathbb{R}^2$ from $(x,f(x))$ to $(y,f(y))$, due to Hausdorff, pathwise connected spaces being arcwise connected (this is Corollary. 31.6 in $\textit{General topology}$, by Willard).

Now calling $\pi_1,\pi_2:\mathbb{R}^2\to\mathbb{R}$ the projections onto the factors, the function $g=\pi_1\circ c:[0,1]\to[x,y]$ would be injective and continuous, and thus an homeomorphism. So $f|_{[x,y]}=\pi_2\circ c\circ g^{-1}:[x,y]\to\mathbb{R}$ would be continuous, contradicting the fact that $f$ is totally discontinuous.

Saúl RM
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