Let $X$ be the rational points of the interval $[0,1] \times 0$ in $\mathbb{R}^2$ and define the point $p = 0 \times 1$. Let $T$ be the union of all line segments joining $p$ with a point of $X$. Show that if $x \in T \setminus \{p\}$, then $X$ is not locally connected at $x$.
I must find a neighborhood $U$ of $x$ such that there is a separation of every neighborhood $V \subseteq U$. The problem is that I cannot seem to find a separation of any neighborhood $U \subseteq T$ of $x$.
Since $V$ is open it will be intersected by infinitely many line segments. The problem is that regardless of how I group the line segments into two sets (to form a separation), I always end up with infinitely many limit point of one set contained in the other. For example, if $A$ is the line segment containing $x$ and $B$ is the union of all other line segment intersecting $U$, then every point of $A$ is a limit point of $B$.
Any help would be greatly appreciated.
Best Answer
Let $x \in T \setminus\{p\}$. If we show that $B_\delta(x) \cap T \setminus\{p\}$ is not connected $\forall 0<\delta <\frac{d(x,p)}{2}$ we have done.
Fixed $\delta \in \left(0,\frac{d(x,p)}{2}\right)$, is easy to see that you can find an irrational number $a$ such that $B_\delta=\{\{B_\delta(x) \cap T \setminus\{p\} \} \cap f^{-1}(-\infty,0)\} \cup \{\{B_\delta(x) \cap T \setminus\{p\}\} \cap f^{-1}(0,\infty)\}$ where the union is disjoint and non-empty, with $f(x,y)=y+\frac{x}{a}-1$. So $B_\delta$ is not connected.
Essentially you want to partition $B_\delta$ in two non-empty open set (relatively to the subspace $T \setminus \{p\}$). $f^{-1}(-\infty,0)$ is the domain "under" the line $y+\frac{x}{a}-1=0$ and $f^{-1}(0,\infty)$ is the "upper" domain. ($f$ is clearly continous so they are open sets)