Prove that $Y$ which has at least cardinality of $\mathbb{R}$, is path connected Let $X$ be an infinite set and $\tau=\lbrace A\subset X : A = \emptyset \quad \text{or} \quad   X \setminus A \quad \text{is  finite}\rbrace$.  Let $Y$ be a subspace of the topological space $(X,\tau)$. Show that if the cardinality of $Y$ is at least the cardinality of 
$\mathbb{R}$, then $Y$ is path connected.
 A: The key to answering this question is to prove that a cofinite or empty subset of $[0,1]$ is open in the usual topology. Given this if $x,y\in Y$ are arbitrary, then since $|Y|\geq |\mathbb{R}|$ there is an injective function $f:[0,1]\to Y$ such that $f(0)=x$ and $f(1)=y$. If $U$ is any non-empty open set in $Y$, then $Y\setminus U$ is finite and so $f^{-1}(U)$ is cofinite in $[0,1]$. Hence by the key step at the start $f^{-1}(U)$ is open in $[0,1]$ and so $f$ is continuous. 
A: Let $\mathbb{I}$ denote the topological space that has $\left[0,1\right]$
as underlying set and is equipped with its usual topology. Let $u,v\in Y$.
A function $p:\mathbb{I}\rightarrow Y$ will be continuous if $p^{-1}\left(A\right)\subset\left[0,1\right]$
is closed for any finite set $A\subset Y$. Consequently every injection
is continuous (then $p^{-1}\left(A\right)$ is also finite, hence closed
in $\mathbb{I}$). If the cardinality of $Y$ is at least the cardinality
of $\left[0,1\right]$ then an injection can be constructed with $p\left(0\right)=u$
and $p\left(1\right)=v$.
