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I am trying to construct a topological space that is path connected, but nowhere locally connected.

If we define $p = (0,1)\in\mathbb{R}^2$, and for all $q\in\mathbb{Q}\cap[0,1]$ $T_q = \{v\in\mathbb{R}^2\ |\ \exists t\in[0,1],\ v = (1-t)p + t(q,0)\}$, then $$ T = \bigcup\limits_{q\in\mathbb{Q}\cap[0,1]}T_q $$ the space of all lines connecting rationals in $[0,1]\times\{0\}$ to $p$, can be shown to be path connected but locally connected only at $p$. I thought of attempts to maybe generalize the idea to make the desired result. The immediate one of these attempts was to maybe 'rational duplicate' $p$ along the $y = 1$ line, i.e. to let $T$ be all the lines connecting every rational point on $[0,1]\times\{0\}$ to every rational point on $[0,1]\times\{1\}$, but the proof attempt did not go (I am not sure if this is correct).

Another idea was maybe a space-filling curve, but this seems a bit exotic.

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1 Answer 1

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In $\mathbb R^2$, draw lines from $(0,0)$ to $(1,q)$ for all rational $q$ in $[0,1]$
and from $(1,0)$ to $(0,-q)$ for all rational $q$ in $[0,1].$ This subspace
is path connected and nowhere locally connected.

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  • $\begingroup$ I think you mean from (0,1) to (-q,0). $\endgroup$ Commented May 6, 2017 at 5:58
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    $\begingroup$ @DanielWainfleet. No, I meant what I said. However, you've brought to my attention an error which I have corrected. $\endgroup$ Commented May 7, 2017 at 2:02
  • $\begingroup$ Nice to see a very simple example $\endgroup$ Commented May 7, 2017 at 2:43
  • $\begingroup$ How should the set be written? I don't have a clear idea, I tried this $\{(x,y)\in\mathbb R^2: xt+(1-t)y\dots\}? $ $\endgroup$
    – user486983
    Commented Jan 3, 2018 at 23:33
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    $\begingroup$ Very nice ... love it! Reference @Isa 's question. WilliamElliot 's description above is very clear, but to "code" it in a way that might seem more rigorous try: $\{(x,y)\in\mathbb{R}^2:y=mx, \text{for some } m \in\mathbb{Q}\cap[0,1] \text{ and some } x\in [0,1]\}$ union $\{(x,y)\in\mathbb{R}^2:y=m(x-1), \text{for some } m \in\mathbb{Q}\cap[0,1] \text{ and some } x\in [0,1]\}$ $\endgroup$ Commented Mar 20, 2018 at 13:42

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