# Path connected subset of $\mathbb R^2$ that is locally connected at none of its points.

Let $X$ denote the rational points of the interval $[0,1]\times\{0\}$ of $\mathbb R^2$. Let $T$ denote the union of all line segments joining the point $p=(0,1)$ to points of $X$.

1. Find a subset of $\mathbb R^2$ that is path connected but is locally connected at none of its points.

My attempts : I got the answer here

I was visualizing the diagram: im not getting how it is Nowhere locally connected ?

Any hints/solution will be appreciated

• I don’t think the diagram is right. You should have lines starting at $(0,0)$ (the origin) and going to $(1,q)$, which means the end points of said lines will all lie on a vertical line with $x$ coordinate equal to $1$. Sep 3 '18 at 13:15

As Clayton observed, the space pictured in your question is not the space $X$ defined in William Elliot's original answer.

Let us redefine $X$ in a slightly modified form. For $r \in \mathbb{R}$ let $L_r$ denote the line segment from $(0,0)$ to $(1,r)$ if $r \ge 0$ and the line segment from $(1,0)$ to $(0,r)$ if $r \le 0$. Now define $$X = \bigcup_{q \in \mathbb{Q}} L_q \subset [0,1] \times \mathbb{R} .$$ Note that the original definition was $X = \bigcup_{q \in [-1,1] \cap \mathbb{Q}} L_q$.

All $L_q$ are path connected subspaces of $X$. Hence those with $q \ge 0$ resp. $q \le 0$ belong to the path component of $(0,0)$ resp. $(0,1)$. But $(0,0), (0,1) \in L_0$ which shows that $X$ is path connected.

Let us check that $X$ is not locally connected at any point $x = (x_1,x_2)\in X$. Define $$V(x) = \begin{cases} X \backslash \{(0,1)\} & x_1 = 0 \\ X \backslash \{(0,0)\} & x_1 = 1 \\ X \backslash \{(0,0), (0,1)\} & 0 < x_1 < 1 \end{cases}$$ This is an open neighborhood of $x$ in $X$. For each $r \in \mathbb{R}$ the set $([0,1] \times \mathbb{R}) \backslash L_r$ splits into two disjoint nonempty open sets $O_r^\pm$ above and below $L_r$. Hence the sets $V_r^\pm(x) = V(x) \cap O_r^\pm$ are disjoint nonempty open subsets of $V(x)$. Call $r$ admissible for $x$ if $r$ is irrational and $r < 0$ for $x_1 = 0$ resp. $r > 0$ for $x_1 = 1$. For an admissible $r$ we have $L_r \cap V(x) = \emptyset$ and therefore $V(x) = V_r^+(x) \cup V_r^-(x)$.

Now consider any open $U \subset X$ such that $x \in U \subset V(x)$. We shall show that there exists a regular $r$ such that both $U_r^\pm = U \cap V_r^\pm(x)$ are nonempty. This proves that $U$ is not connected.

Since $U$ is open, the set $S(U) = \{ q \in \mathbb{Q} \mid L_q \cap U \ne \emptyset \}$ is easily seen to be open in $\mathbb{Q}$.

a) $x_1 = 0$. Then $x = (0,q)$ with $q \le 0$. Obviously $q\in S(U)$. Choose $p \in S(U)$ such that $p < q$ and an irrational $r$ such that $p < r < q$. Then $r$ is admissible for $x$ and both $U_r^\pm$ are nonempty.

b) $x_1 = 1$. Can be treated similarly.

c) $0 < x_1 <1$. Choose $p, q \in S(U)$ such that $p < q$ and an irrational $r$ such that $p < r < q$. Then $r$ is admissible for $x$ and both $U_r^\pm$ are nonempty.