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Define a set X $:=\mathbb{Z}\times\{0,1\}$ and declare $\{(-n,0)\},\{(-n,1)\},\{(0,0)\},\{(0,0),(0,1)\},\{(n,0),(n,1)\}$as open subsets of X. This forms the basis for a topology on X.

All sets C of the topology are finite except X itself. And since X is countably infinite, X\C are finite. And I do not any have argument against the fact that the topology is Co-finite (I have no complements that are open).

I also know that all Co-finite topologies on infinite set are connected. This is intuitively apparent since for it to be disconnected, there would need to be two open sets whose disjoint union is the original topology. (Correct me if I am incorrect).

Can I conclude from this that the topological space defined by the basis above is connected? What about path connectedness? Is more information needed?

Note, However, I do not have firm grasp of Bases and Relation to Topological which could be possible pitfall in my argument.

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I have removed the tag path-connectedness. To me this tag seems to specialized. The tag wiki for connectedness says, that it is intended for questions concerning path-connectedness, too. If you think that we should keep this tag, feel free to bring this up in tagging chatroom or on meta. – Martin Sleziak Apr 15 '13 at 9:50
up vote 1 down vote accepted

Well, the topology itself usually means the set of all open subsets, and open subsets are generated by union of arbitrarily many basis sets.

So it's not true, that every set of the topology (i.e. every open set) is finite, because for example $$C:=\{(-1,0),(-2,0),(-3,0),(-4,0),\dots\}$$ is open in $X$. This also gives a counterexample that $X\setminus C$ can be infinite, but it can be finite as well for some larger $C$.

And, $X$ is not connected, for example $X\setminus\{(-1,0)\}$ can be written as union of basis sets, so it is open, but $\{(-1,0)\}$ is also open. Consequently, $X$ cannot be neither path connected. (Anyway, forget about path connectedness in a countable space!)

About the co-finite topologies (where all open sets, except for $\emptyset$ are co-finite, i.e. their complements, all closed sets are finite), it is indeed obvious that such must be connected, but this is irrelevant here.

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Thanks a lot. I had overlooked a few details. – user45099 Apr 8 '13 at 22:50
Concerning "Anyway, forget about path connectedness in a countable space!", I know that holds for $T_1$ spaces. I know an indiscrete space is path-connected. Is there a weaker condition than $T_1$ that implies a countable space with such a topology cannot be path-connected? – Daniel Fischer Dec 19 '13 at 21:52

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