Is the real number line $\mathbb{R}$ with the compact complement topology simply connected?

Question

Let $\mathbb{E}^1$ denote $\mathbb{R}$ equipped with the standard Euclidean topology. Let $X$ denote $\mathbb{R}$ with the topology for which a nonempty subset $U \subset \mathbb{R}$ is open if and only if its complement $\mathbb{R} \backslash U$ is compact as a subspace of $\mathbb{E}^1$. This is called the compact complement topology in [SS].

Question: Is $X$ simply connected?

To paint a bit of a picture, note that the topology on $X$ is coarser than the Euclidean topology. This means that the identity function $\mathbb{E}^1 \to X$ is continuous, so it clearly follows that $X$ is arc-connected (each pair of distinct points is connected by an injective path $[0, 1] \to X$). The topology on $X$ is finer than the finite complement topology on $\mathbb{R}$, for which nonempty open sets are exactly the complements of finite sets; this is equivalent to $X$ being $T_1$ [SS, Item 1]. Observe also that $X$ has no disjoint nonempty open subsets, which means it is hyperconnected (= irreducible) [SS, Item 3]. It easily follows from hyperconnectedness that $X$ is not Hausdorff. Unlikely, but perhaps it's relevant to mention that $X$ is compact, since the complement of any chosen set in an open cover is compact, so finitely many of the others plus the chosen set must cover $X$ [SS, Item 4].

This answer in the thread Path-connected subsets of compact complement topology says that any subset of $X$ containing an unbounded interval is path-connected, and speculates that a set which does not contain an unbounded interval is path-connected if and only if it is path-connected as a subspace of $\mathbb{E}^1$. Taking this for granted, it seems like the idea may be that the path $[0, 1] \to X$ from $0$ to $1$ which is the restriction of the identity function shouldn't be path homotopic to a path from $0$ to $1$ with image $(-\infty, 0] \cup \{1\}$, but I don't know how to argue this.

The thread Is the compact complement topology locally arc connected? may also have potentially useful information.

It may be interesting to compare with superficially similar spaces. I.e,

• The finite complement topology on $\mathbb{R}$ is contractible. See here.
• The countable complement topology on $\mathbb{R}$, for which open sets are the empty set and the complements of countable sets, is connected but not even path-connected. See this answer.

[SS] Steen, L. A.; Seebach, J. A. jun., Counterexamples in topology, New York etc.: Holt, Rinehart and Winston, Inc., XIII, 210 p. (1970). ZBL0211.54401.

Answer 1

The answer is Yes.

Definition. Let $U(x,\epsilon)=(-\infty,x-\epsilon^{-1})\bigcup (x-\epsilon,x+\epsilon)\bigcup (x+\epsilon^{-1},\infty)$ be some special open set in $X$.

Note that a function $f$ from an arbitrary space $Y$ to $X$ is continuous iff: when identifying $X$ as $\mathbb{R}$, for any point $p\in Y$ and any $\epsilon>0$, there is a neighborhood of $p$ which for every point $q$, $f(q)\in U(f(p),\epsilon)$.

Definition. For $p,q\in X$, "The straight line" from $p$ to $q$ is a path: $f(t)‎ = p+t(q-p),t\in [0,1]$. We denote $f$ as $L(p,q)$.

Definition. Let $K_+ (p,q)$ be the path: $f(t)‎ = \frac{t}{1-t}+p, t\in [0,1)$ and $f(1)=q$. Let $K_- (p,q)$ be the path: $f(t)‎ = \frac{t}{t-1}+p,t\in [0,1)$ and $f(1)=q$. One can verify these two paths are continuous (as map to $X$).

Claim 1. $K_+(0,0)$ is null-homotopic. As a corollary, $K_+(p,q)$ and $K_- (p,q)$ are homotopic to $L(p,q)$.

Proof. Let $f=K_+(0,0)$. We construct another loop $g:[0,1]\rightarrow X$, which is "many $f$ piled up": For $x\in [1-\frac{1}{2^{n-1}},1-\frac{1}{2^{n}}) (n\in \mathbb{Z}^+)$, say $x=(1-t)(1-\frac{1}{2^{n-1}}) + t(1-\frac{1}{2^{n}})$, let $g(x)=f(t)+n-1$, and let $g(1)=0$. One can use the criterion above to verify that $g$ is continuous. Let $h_1:[-1,1]\rightarrow X$, $h_1(x)=f(x+1)$ when $x<0$, $h_1(x)=g(x)+1$ when $0\le x<1$ and $h_1(1)=0$. We can view $h_1$ as the product of $f$ and $g$. Let $h_2(x)=x+1$ when $-1\le x<0$ and $h_2(x)=h_1(x)$ when $x\ge 0$. Obviously $g$ and $h_2$ are homotopic. Now we construct a homotopy $H: [-1,1]\times [0,1] \rightarrow X$ from $h_1$ to $h_2$: \begin{equation} H(s,t)=\begin{cases} s+1, & \text{if $s<-t$}\\ h_1(\frac{s+t}{1+t})-t, &\text{if $-t\le s<1$}\\ 0, &\text{$s=1$} \end{cases} \end{equation} One can verify that $H$ is continuous. So the loops $h_1$ and $h_2$ are homotopic, hence $g\simeq f*g$, hence $f\simeq 0$. The corollary is trivial.

Claim 2. Every loop $f: [0,1]\rightarrow X$, which $f(0)=f(1)=q$ and $f(x) \in U(q,1)$ for every $x$, is null-homotopic.

Proof. The proof is essentially the same as the previous one. We can assume $q=0$. For $s,t\le 1$, let $$ \phi_{s,t}(x)=\begin{cases} ts^{-1} x, & |x|<s\\ st^{-1} x, & |x|>s^{-1} \end{cases} $$ be a continuous one-to-one mapping from $U(0,s)$ to $U(0,t)$. The loop $g:[0,1]\rightarrow X$: for $x\in [1-\frac{1}{2^{n-1}},1-\frac{1}{2^{n}}) (n\in \mathbb{Z}^+)$, say $x=(1-t)(1-\frac{1}{2^{n-1}}) + t(1-\frac{1}{2^{n}})$, let $g(x)=\phi_{1,(n+1)^{-1}} \circ f(t)$, and $g(1)=0$. As the previous case, extend $g$ to $h_2$: $h_2(x)=0$ for $-1\le x<0$ and $h_2(x)=g(x)$ for $0\le x\le 1$. Let $h_1$ be the homotopy product: $h_1(x)=f(x+1)$ for $-1\le x<0$ and $h_1(x)=g(x)$ for $0\le x\le 1$. The homotopy is given by \begin{equation} H(s,t)=\begin{cases} 0, & \text{if $s<-t$}\\ \phi_{1,(n+1-t)^{-1}}\circ f(w), &\text{if $-t\le s<1$ and $\frac{s+t}{1+t}=(1-w)(1-\frac{1}{2^{n-1}}) + w(1-\frac{1}{2^{n}})$}\\ 0, &\text{$s=1$} \end{cases} \end{equation} Hence $f\simeq 0$.

Finally we solve the original problem. It suffices to prove that any loop $f$ which $f(0)=f(1)=0$ is null-homotopic.

By continuity, for every $e\in [0,1]$, there is a neighborhood (open interval containing $e$) where $f(x)\in U(f(e),1)$ for each $x$. By the compactness of $[0,1]$, it is covered by finite number of those intervals, so we can divide $[0,1]$ into $n$ parts: $[0(=d_0),d_1],\dots,[d_{n-1},d_n(=1)]$, such that for each interval there exists $e_i\in [d_{i-1},d_i]$ that $f(x)\in U(f(e_i),1)$ for $x$ in $[d_{i-1},d_i]$. Let $f_i$ be the path induced by this interval. We will show that $f_i$ is homotopic to $L(f(d_{i-1}),f(d_i))$, hence the whole loop is null-homotopic.

Let $a=L(f(e_i),f(d_{i-1})), b=L(f(d_i),f(e_i))$. We show that there exists a path $c$ from $f(e_i)$ to $f(d_{i-1})$ with value in $U(f(e_i),1)$ which is homotopic to $a$. Indeed, if $d_{i-1}$ lies in $(f(e_i)-1,f(e_i)+1)$, let $c=a$; if it lies in $(f(e_i)+1,\infty)$, let $c=(K_+ (f(d_{i-1}),f(e_i)))^{-1}$ be the reverse path of $K_+$. By claim 1, $c$ and $a$ are homotopic. If $d_{i-1}$ lies in $(-\infty,f(e_i)-1)$, let $c= (K_- (f(d_{i-1}),f(e_i)))^{-1}$.

In the same way, let $d$ be the path from $f(d_i)$ to $f(e_i)$ with value in $U(f(e_i),1)$ and homotopic to $b$. By our second claim, $c* f_i * d$ (as a loop valued in $U(f(e_i),1)$) is a null-homotopic loop, hence $f_i$ is homotopic to $a^{-1} * b^{-1}$, hence homotopic to the straight line $L(f(d_{i-1}),f(d_i))$ . The proof is done.