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Let $\tau$ be the standard topology on $\mathbb{R}$ and let $\tau'$ be the compact complement topology on $\mathbb{R}$: $$\tau' = \{\mathbb{R} \setminus K \mid K \text{ is $\tau$-compact}\} \cup \{\varnothing, \mathbb{R}\}$$

π-Base claims that $(\mathbb{R},\tau')$ is locally arc-connected. I believe the proof that it is arc-connected: since $\tau' \subseteq \tau$, the identity map $id:(\mathbb{R}, \tau) \rightarrow (\mathbb{R},\tau')$ is a continuous injection and thus gives an arc between any two distinct points.

They give the same proof for local arc-connectedness, but it is insufficient since we must show that every nonempty open set contains an arc-connected open neighborhood of each of its points. In particular, I'm skeptical that any nonempty, proper, $\tau'$-open set can even be path-connected.

Steven Clontz
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  • Isn't every $\tau'$-open subset homeomorphic to $(X,\tau')$? – Henno Brandsma May 08 '17 at 21:06
  • @HennoBrandsma Do you have a proof of this? –  May 08 '17 at 21:57
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    In fact, no proper subset that is unbounded on both sides is homeomorphic to $(\mathbb{R},\tau')$. The space $(\mathbb{R},\tau')$ has the property that every Hausdorff subset is contained in a connected Hausdorff subset, but no proper subset that is unbounded on both sides has the same property. (Note that a subset is Hausdorff iff it is bounded.) – Eric Wofsey May 08 '17 at 22:07
  • @EricWofsey I see that a subset is Hausdorff iff it is bounded and thus that every Hausdorff subset is contained in the connected Hausdorff subset $[-N,N]$ for some $N>0$. Why do the sets unbounded on both sides fail to satisfy this property? –  May 08 '17 at 22:40
  • If $X\subset\mathbb{R}$ is a proper subset that is unbounded on both sides, let $a\in\mathbb{R}\setminus X$ and choose $x,y\in X$ with $x<a<y$. Then there is no connected Hausdorff subset of $X$ containing ${x,y}$. (So you don't need much from "unbounded": you just need to contain points on both sides of some $a\in\mathbb{R}\setminus X$.) – Eric Wofsey May 08 '17 at 22:42
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    Actually, by ad hoc arguments, you can rule out all the other cases and show that no proper subset of $(\mathbb{R},\tau')$ is homeomorphic to it. If no such $x<a<y$ exists and $X$ is compact but not Hausdorff, then $X$ must be a closed unbounded interval; say $X=[x,\infty)$. But then you can remove one point from $X$ to obtain $(x,\infty)$, which has the property that every Hausdorff subset is contained in a connected Hausdorff subset. Since you can't remove a point from $\mathbb{R}$ to get a subset with that property, $\mathbb{R}\not\cong[x,\infty)$. – Eric Wofsey May 08 '17 at 22:48

1 Answers1

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You are correct that the given argument does not work. However, $(\mathbb{R},\tau')$ is in fact locally arc-connected. It suffices to show that if $a<b<c<d$ then the open set $U=(-\infty,a)\cup(b,c)\cup(d,\infty)$ is arc-connected, since such open sets form a basis. I will show that if $x\in (d,\infty)$ and $y\in (b,c)$ then there is a continuous injection $f:[0,1]\to U$ with $f(0)=x$ and $f(1)=y$; the other cases are similar.

To define $f$, first let $g:[0,1)\to[x,\infty)$ be an order-isomorphism. Then define $f(t)=g(t)$ for $t\in[0,1)$ and $f(1)=y$. Clearly $f$ is injective, and it is continuous at any $t\in[0,1)$ since $g$ is continuous with respect $\tau$ and hence also with respect to $\tau'$. So we just need to check continuity at $1$. If $V\subseteq U$ is any $\tau'$-open set containing $y$, then $V$ must contain $(z,\infty)$ for some $z\in[x,\infty)$. Thus $f^{-1}(V)$ contains the entire interval $(g^{-1}(z),1]$, and in particular contains a neighborhood of $1$. Thus $f$ is continuous at $1$.

Eric Wofsey
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