Is a set with the cofinite topology path-wise connected? Let $X$ be a general set. I wonder if you can state if it is not path-wise connected regardless what set it is. I guess it would depend on the cardinality of $X$. $$ $$
If $X$ is finite, and has 2 or more points it is not connected, therefore cannot be path-wise connected. (Since path-wise connectedness implies connectedness.) I have found a proof which shows $\mathbb{N}$ is not path-wise connected with this topology.(Prove that $\mathbb{N}$ with cofinite topology is not path-connected space.) Does this hold true if $X$ is countable in general? $$ $$And what about sets with other cardinality?
 A: Let $X$ be a topological space equipped with cofinite topology, and such that its cardinality equals or exceeds the cardinality of $[0,1]$.
Then an injection $f:[0,1]\to X$ exists. 
Now let $a,b\in X$ and prescribe $\gamma:[0,1]\to X$ by $0\mapsto a$, $1\mapsto b$ and $t\mapsto f(t)$ for $t\in(0,1)$. 
Note that preimages of finite subsets of $X$ under $\gamma$ are finite, hence are closed. 
That implies here that $\gamma$ is a continuous function. 
Proved is now that $X$ is path-connected.
A: A result that I think is from Sierpinski is that $[0,1]$ is not the union of a countable family $F$ of pair-wise disjoint non-empty closed subsets unless $F=\{[0,1]\}.$
If $X$ is countably infinite and has the co-finite topology, then $X$ is a $T_1$ space.
Suppose $f:[0,1]\to X$ is continuous with $f(0)\ne f(1).$ Then $F=\{f^{-1}\{x\}: x\in X\}$ \ $\{\phi\}$ is a countable family of pair-wise disjoint non-empty closed subsets of $[0,1]$ and $F$ has  more than $1$ member, which is impossible. So every continuous $f:[0,1]\to X$ is constant, and $X$ is not path-connected. 
