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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 $\mathbb R^{\mathbb R}$?
Asaf Karagila
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Snared
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1 Answers1

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The previously accepted answer has an oversight. Consider any function $\vec x\in\mathbb R^{\mathbb R}\setminus\{\vec 0\}$ where we have $\vec x(0)=1$ (to prove $\vec x\not=\vec 0$) and $\vec x(1)=0$. Then $\vec x(1)\not\in A_{1}^+=\mathbb R\setminus\{0\}$ and $\vec x(1)\not\in A_{1}^-=\mathbb R\setminus\{0\}$. Therefore the claim $A^+\cup A^-=\mathbb R^{\mathbb R}\setminus\{\vec 0\}$ is false.

Indeed, let $\vec z\in \mathbb R^{\mathbb R}$; we will show that $\mathbb R^{\mathbb R}\setminus\{\vec z\}$ is path connected. Let $\vec x,\vec y\in\mathbb R^{\mathbb R}\setminus\{\vec z\}$. Choose $a_x$ with $\vec x(a_x)\not=\vec z(a_x)$ and choose $a_y$ with $\vec y(a_y)\not=\vec z(a_y)$. We define a path $p:[0,1]\to\mathbb R^{\mathbb R}$ from $\vec x$ to $\vec y$ by $$ p(t)(a) = \begin{cases} (1-t)\vec x(a)+t\vec y(a) & \text{if }a\not\in\{a_x,a_y\} \\ \vec x(a_x) & \text{if }a=a_x, t\leq\frac{1}{2} \\ (2-2t)\vec x(a_x)+(2t-1)\vec y(a_x) & \text{if }a=a_x, t\geq\frac{1}{2} \\ (1-2t)\vec x(a_y)+2t\vec y(a_y) & \text{if }a=a_y, t\leq\frac{1}{2} \\ \vec y(a_y) & \text{if }a=a_y, t\geq\frac{1}{2} \\ \end{cases} $$ which is continuous as it is continuous on every coordinate. Furthermore, note that for $t\leq\frac{1}{2}$ we have $p(t)(a_x)=\vec x(a_x)\not=\vec z(a_x)$ and for $t\geq\frac{1}{2}$ we have $p(t)(a_y)=\vec y(a_y)\not=\vec z(a_y)$, so in general we have $p(t)\not=\vec z$.

There's a small problem though; what if $a_x=a_y$? Then the above path is not well-defined, but we may use this path instead $$ p(t)(a) = \begin{cases} (1-t)\vec x(a)+t\vec y(a) & \text{if }a\not\in\{a_x,a_x+1\} \\ \vec x(a_x) & \text{if }a=a_x, t\leq\frac{1}{3} \\ (2-3t)\vec x(a_x)+(3t-1)\vec y(a_x) & \text{if }a=a_x, \frac{1}{3}\leq t\leq\frac{2}{3} \\ \vec y(a_x) & \text{if }a=a_x, t\geq\frac{2}{3} \\ (1-3t)\vec x(a_x+1)+3t(\vec z(a_x+1)+1) & \text{if }a=a_x+1, t\leq\frac{1}{3} \\ \vec z(a_x+1)+1 & \text{if }a=a_x+1, \frac{1}{3}\leq t\leq\frac{2}{3} \\ (3-3t)(\vec z(a_x+1)+1)+(3t-2)y(a_x+1) & \text{if }a=a_x+1, t\geq\frac{2}{3} \\ \end{cases} $$ noting that for $t\leq \frac{1}{3}$ we have $p(t)(a_x)=\vec x(a_x)\not=\vec z(a_x)$, for $\frac{1}{3}\leq t\leq \frac{2}{3}$ we have $p(t)(a_x+1)=\vec z(a_x+1)+1\not=\vec z(a_x+1)$ , and for $t\geq \frac{2}{3}$ we have $p(t)(a_x)=\vec y(a_x)\not=\vec z(a_x)$, so in general we have $p(t)\not=\vec z$.

Steven Clontz
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