2
$\begingroup$

There are abundant counterexamples in literature of the $2$ statements -

  1. $X$ is Path Connected $\implies$ $X$ is Locally Path Connected
  2. $X$ is Arc Connected $\implies$ $X$ is Locally Arc Connected

In all of the counterexamples I've found, they hold as the space is Path/Arc Connected, but is not Locally Connected (for example, the Extended Topologist's Sine Curve and the Closed Infinite Broom).

So, I wish to ask - are there counterexamples of the above $2$ statements, if we also assume that $X$ is Locally Connected? How about Locally Path Connected for Statement $2$?

$\endgroup$
1
  • 1
    $\begingroup$ Note that a space that is both locally path-connected and connected space is path-connected. So we don't have an example for the "mixed" case. $\endgroup$ Aug 5, 2020 at 4:42

1 Answer 1

1
$\begingroup$

Let $Y=[0,1]\times[0,1]$ with the lexicographic order topology. For each $x\in[0,1]$ let $I_x$ and $I^x$ be copies of $[0,1]$ with its usual topology, and for each $t\in[0,1]$ let $t_x$ and $t^x$ be the copies of $t$ in $I_x$ and $I^x$, respectively. For $x\in[0,1]$ identify $\langle x,0\rangle\in Y$ with $0_x\in I_x$ and $\langle x,1\rangle\in Y$ with $0^x\in I^x$. Then identify all of the points $1_x$ and $1^x$ to a single point $p$ to get the space $X$.

Informally, we attach a ‘sticker’ in the form of a copy of the closed unit interval to each point on the bottom and top edges of the lexicographically ordered square, and we identify the free ends of the stickers.

Then $X$ is path connected and locally connected, but it is not locally path connected at any of the points $\langle x,0\rangle\sim 0^x$ or $\langle x,1\rangle\sim 0^x$.

$\endgroup$
4
  • $\begingroup$ What exactly do you mean by $\langle x,0\rangle\sim 1^x$? $\endgroup$
    – Ishan Deo
    Aug 6, 2020 at 6:02
  • 1
    $\begingroup$ @IshanDeo: Argh! It was supposed to be $\langle x,0\rangle\sim 0^x$, the point of $X$ that results from the identification of $\langle x,0\rangle\in Y$ and $0^x\in I^x$. I don’t know whether it was a typo or a mental hiccup, but I’m fixing it as soon as I post this comment! $\endgroup$ Aug 6, 2020 at 7:14
  • $\begingroup$ So, correct me if I'm wrong, but - you added the lines $I_x$ and $I^x$ for the sole purpose of creating a path between any $2$ points of $X$, which did not originally exist in the ordered square. Also, $X$ is not locally path connected for the same reason the ordered square is not. $\endgroup$
    – Ishan Deo
    Aug 6, 2020 at 8:22
  • $\begingroup$ @IshanDeo: That’s right, on both counts. $\endgroup$ Aug 6, 2020 at 8:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .