Consider $[0,1]$ with the co-countable topology. This means that the open sets are the empty set $\emptyset$ and all $A\subset [0,1]$ such that $[0,1]\setminus A$ is a countable (or finite) set.

Consider the following argument:

*Take any sequence $(a_n)_{n\geq 1}$ contained in $[0,1]$ such that $\,a_n$ converges to $a\in [0,1]$.* *Since $\,([0,1]\setminus \{a_n:n\geq 1\}) \cup \{a\}$ is a neighborhood of $\,a$, we conclude that it must be that the sequence $\,(a_n)_{n\geq 1}$ is eventually constant, and equal to $\,a$.*

*Let $\,f:[0,1]\to \mathbb{R}$ be a real valued function from the above topological space to $\mathbb{R}$ equipped with the usual Euclidean topology. More specifically, let us take $\,f(x)=1$ for $\,x\in [0,1/2]$, and $\,f(x)=2-2x$ for $\,x\in[1/2,1]$.*

*The just made observation about the convergence of sequences on $\,[0,1]$ implies that if $\,\lim_n a_n =a$ then $\,\lim_n f(a_n) = f(a)$. However, $\,f^{-1}(\{1\})= [0,1/2]$ is not a closed set in$\, [0,1]$ with co-countable topology, and so $\,f$ cannot be a continuous function.*

Which of the following claims are **TRUE**?