Is $\mathbb{R}$ under the countable complement topology path connected? (Proof check)

Question

I'm trying to prove that when $\mathbb{R}$ has the countable complement topology, it is not path connected.

I used the following definition of continuous: $f(\overline{A})\subset\overline{f(A)}$.

We want to show that some path $f: [0,1] \rightarrow{} \mathbb{R}$ cannot be continuous. Taking $A=\mathbb{Q} \cap [0,1]$, we get that $f(\overline{A})=f([0,1])$, which is uncountable (by density of rationals within the reals). However, $f(A)$, which is countable, has no limit points in the countable complement topology (for each other point, one can construct a neighborhood that avoids exactly the countably many points in $f(A)$). This means that $\overline{f(A)}=f(A)$ is also countable. However, an uncountable set cannot be a subset of a countable set.

Does this proof seem sufficient?

Answer 1

I'm not sure what you mean by "density of rationals within the reals" - the reals with this topology are not separable, and thus the rationals cannot be a dense subset. (In fact, the rationals are closed as they are countable.)

This is S17 of the pi-Base, and example #20 in Steen/Seebach's Counterexamples in Topology. The pi-Base deduces this space is not path connected based on the fact that all anticompact $T_1$ spaces are totally path disconnected.

To see this, let $f:[0,1]\to\mathbb R$ be a path. Then $f([0,1])$ must be compact and connected as the continuous image of a compact and connected space. Since $\mathbb R$ only has finite sets as compact sets, and the only finite connected $T_1$ space is the singleton, every path is a single point.

To see that your space is $T_1$, note $\mathbb R\setminus\{x\}$ is open and thus singletons are closed. To see that your space is anticompact, let $Y\subseteq\mathbb R$ be infinite, and let $\{y_n:n<\omega\}\subseteq Y$ be distinct. Then $\{\mathbb R\setminus\{y_n:N\leq n<\omega\}:N<\omega\}$ is an open cover of $Y$ with no finite subcover, showing it is not compact.

Answer 2

The claim that the image of $[0,1]$ has to be uncountable is incorrect. In fact, given the anti-discrete topology on some set $Y$, and a subset $B\subseteq Y$ we can always find a continuous function $[0,1]\to Y$ with $B$ as its image (obviously as long as $|B|\leq \mathfrak{c}$). And so, in general a continuous image of $[0,1]$ can be of any cardinality up to $\mathfrak{c}$. And unfortunately I don't see a way to fix your proof, it is a fatal flaw.

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But the statement is still true: the cocountable topology is not path-connected. In fact it is totally path-disconnected. And it follows from the following facts:

1. Compact subsets of cocountable topology are precisely the finite subsets.
2. Points are closed in the cocountable topology, and so finite subsets are discrete, in particular totally disconnected.
3. Continuous image of compact is compact. And of connected is connected.
4. $[0,1]$ is both compact and connected.

Putting everything together we get that a continuous function $[0,1]\to X$, with cocountable topology on $X$, has to be constant. And therefore $X$ is not path-connected, as long as it has more than $1$ point.

Answer 3

I think your proof is not finished: Let $B=f([0,1])$. You just showed that $B$ is countable. Now, take a point $y\in B$. Then $C_1=\{y\}$ and $C_2=B/\{y\}$ are closed subsets. Since $f$ is continuous, $f^{-1}(C_1)$ and $f^{-1}(C_2)$are closed. But, $f^{-1}(C_1)\cap f^{-1}(C_2)=\emptyset$ and $[0,1]$ is connected. So, $C_2=\emptyset$ and $f$ is constant map.