Vtyjkyuk

Let $X$ denote the rational points of the interval $[0,1]times0$ of $mathbb R^2$. Let $T$ denote the union of all line segments joining the point $p=(0,1)$ to points of $X$.

2. 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

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 $r in mathbbR$ 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
$$X = bigcup_q in mathbbQ L_q subset [0,1] times mathbbR .$$
Note that the original definition was $X = bigcup_q in [-1,1] cap mathbbQ L_q$.

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
$$V(x) =
begincases
X backslash (0,1) & x_1 = 0 \
X backslash (0,0) & x_1 = 1 \
X backslash (0,0), (0,1) & 0 < x_1 < 1
endcases
$$
This is an open neighborhood of $x$ in $X$. For each $r in mathbbR$ the set $([0,1] times mathbbR) backslash L_r$ 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 mathbbQ mid L_q cap U ne emptyset $ is easily seen to be open in $mathbbQ$.

a) $x_1 = 0$. Then $x = (0,q)$ with $q le 0$. Obviously $qin S(U)$. 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.