Is $f:\mathbb E^1\to X$ continuous? $f(x)=x$.
$X$ is the set of all real numbers with finite complement topology (A set is open in this space iff it's complement is finite).
 A: This follows from a general fact: If $\mathcal{O}$, $\mathcal{O}^\prime$ are two topologies on a set $X$ such that $\mathcal{O}$ is finer than $\mathcal{O}^\prime$ (i.e., every set in $\mathcal{O}^\prime$ is also in $\mathcal{O}$), then the identity function $\mathrm{id}_X : x \mapsto x$ is continuous as a function from $\langle X , \mathcal{O} \rangle$ to $\langle X , \mathcal{O}^\prime \rangle$.  (In fact, the converse also holds: if the identity function $\mathrm{id}_X : x \mapsto x$ is continuous as a function from $\langle X , \mathcal{O} \rangle$ to $\langle X , \mathcal{O}^\prime \rangle$, then the topology $\mathcal{O}$ is finer than the topology $\mathcal{O}^\prime$.)
A: $f$ is continuous since $\mathbb R \setminus$ (any finite set) is open in the standard space  $\mathbb R$.
Let $F$ be a finite subset of $\mathbb R$, and let {$x_1,...,x_n$} be the increasing enumeration of $F$.  Then $f^{-1} (\mathbb R \setminus F)=(-\infty, x_1)\cup (x_1,x_2)\cup ... \cup (x_n,\infty)$ is open.
Edit: Alternatively you could just notice the closed sets in $X$ are the finite sets, which are also closed in the standard $\mathbb R$.
