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

I want to show that $T$ is locally connected only at the point $p$. I have been thinking for hours but can't come up with a way. Help would be appreciated.

  • $\begingroup$ Which part are you having trouble with, local connectedness at $p$ or failure of local connectedness at other points? $\endgroup$
    – Lee Mosher
    Feb 13 '17 at 21:32
  • $\begingroup$ @LeeMosher Mainly the latter, but an explanation of both would be helpful $\endgroup$ Feb 13 '17 at 21:34

The local connectedness at $p$: $\Big(\,B(p, \frac{1}{n}) \bigcap T\,\Big)_n$ clearly is a connected neighborhood basis at $p$ in $T$.

Let $y \in T - \{p\}$. Then any neighborhood $U$ of $y$ in $T$ such that $p\not\in U$ doesn't contain any open connected neighborhood of $y$, as given $\epsilon < |y-p|$, $B(y,\epsilon) \bigcap T$ necessarily contains two-by-two disjoint lines

  • $\begingroup$ how $B(y,\epsilon) \bigcap T$ necessarily contains two-by-two disjoint lines? can u elaborate more $\endgroup$
    – Lily
    Apr 12 '18 at 6:26

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