Let $U \subset \mathbb R^{\mathbb R}$. We are looking for a minimal cardinality $U$ where $\mathbb R^{\mathbb R} - U$ is not path connected.
First of all, it is obvious that there will be an injection $U \to \mathbb R^k$ for any $k>0$, since we can always find an imbedding $f : \mathbb R^k \to \mathbb R^{\mathbb R}$, take the image of $f$, and "go around" whichever coordinates are missed (Say, a line using dimension $k+1$, where we never move across the $k$ dimensions).
My guess is that $U$ will have to be of equal cardinality to $\mathbb R^{\mathbb R}$ - since an infinite sphere with $\mathbb R$ dimensions almost seems necessary to remove from $\mathbb R^{\mathbb R}$ to disconnect $\mathbb R^{\mathbb R}$.
My question is:
- Is there a smaller set that disconnects $\mathbb R^{\mathbb R}$?
- If not, is that canonical sphere the same size as $\mathbb R^{\mathbb R}$?
Let $A^+ = \prod_{t \in ℝ} A^+_t$, where $$ A^+_t= \begin{cases} (0,∞), & t=0, \\ ℝ - 0, & t \neq 0. \end{cases} $$ Define $A^-$ analogously with $A^-_0 = (-∞, 0)$.
Sets $A^-$, $A^+$ are non-empty, disjoint, and open in $A^+ \cup A^- = ℝ^ℝ - \{0\}^ℝ$. Thus this space is disconnected. Therefore, the singleton $U = \{\{0\}^ℝ\}$ will suffice to disconnect $ℝ^ℝ$ (as William M. suggested in the comments). Disconnected space cannot be path-connected, so the proof is finished.
Alternative answer: Use the contrapositive of the main theorem of connectedness, the fact that any projection $π:ℝ^ℝ−\{0\}^ℝ →ℝ-0$ is continuous in natural topology, and the fact that image of projection, i.e. $ℝ-0$, is not connected.