Prove that if $f:\mathbb{R}\to\mathbb{R}$ is continuous, then it is continuous from the right I'm trying to prove that if $f:\mathbb{R}\to\mathbb{R}$ is continuous (where the topology of $\mathbb{R}$ is $\emptyset$, $\mathbb{R}$, and all sets of the form $(-\infty, a)$), then it is continuous from the right (i.e. $\forall x\in \mathbb{R}$ and $\epsilon >0, \exists \delta>0$ s.t. $x\leq x' <x+\delta$ implies $|f(x)-f(x')|<\epsilon$. This is what I have so far: 
Let $|f(x)-f(x')|<\epsilon$. That is $f(x)-f(x')\in (-\epsilon, \epsilon)\subset [-\epsilon, \epsilon]$. Since $f$ is continuous, the preimage of a closed set is also closed, so $\exists \delta > 0$ s.t. $f^{-1}([-\epsilon, \epsilon]) = [-\delta, \delta]$, so $f^{-1}(f(x)-f(x'))\in [-\delta, \delta]$. I'm not sure if I'm on the right track, but I can't get the inequalities to work out. Also, does preimage distribute in the sense $f^{-1}(f(x)-f(x'))=x-x'$?
 A: The condition of continuity for the topology you described is as follows:

$\newcommand{\reals}{\mathbb{R}}f : \reals \to \reals$ is continuous iff for each $x \in \reals$ and each $\epsilon > 0$ there is a $\delta > 0$ such that $f(y) < f(x) + \epsilon$ for all $y < x + \delta$.

(Very brief sketch: If $f$ is continuous, given $x \in \reals$ and $\epsilon > 0$ note that $f^{-1} [\;(-\infty,f(x)+\epsilon)\;]$ is an open neighbourhood of $x$, so it is either $\reals$ or of the form $(-\infty , a )$ for $a > x$.  If $f$ satisfies the condition, then given any open $V = ( -\infty,b ) \subseteq \reals$ if $f^{-1}[V] \neq \varnothing , \reals$ then any element of $\reals \setminus f^{-1}[V]$ can be shown to be an upper bound of $f^{-1}[V]$, so $a := \sup f^{-1}[V]$ exists, and we can show that $f^{-1} [V] = (-\infty,a)$.)
Now recall the $\epsilon$-$\delta$-definition of continuous from the right:

$f : \reals \to \reals$ is continuous from the right iff for each $x \in \reals$ and each $\epsilon > 0$ there is a $\delta > 0 $ such that $| f(y) - f(x) | < \epsilon$ for all $x \leq y < x+\delta$.

Note that if $f$ is non-decreasing the $\epsilon$-$\delta$-condition above simplifies somewhat to

for each $x \in \reals$ and each $\epsilon > 0$ there is a $\delta > 0$ such that $f(x) \leq f(y) < f(x) + \epsilon$ for all $x \leq y < x + \delta$.

Given a continuous $f : \reals \to \reals$, as you have already proved, $f$ is non-decreasing, so it is pretty much a matter of comparing the above to the condition of continuity.
