Show that $f$ is partly continuous when $f$ is continuous with regards to the topology formed by the intervals $(a, \infty)$, where $a \in \Bbb R$. 
Let $X$ be a space. A map $f : X \to \Bbb R$ is downward partly continuous if for all $a \in X$ and for every $\varepsilon >0$ there is a open set $U_a$ such that $f(x) > f(a) - \varepsilon$ for all $x \in U_a$. Show that $f$ is partly continuous when $f$ is continuous  with regards to the topology $\tau$ formed by $\emptyset, \Bbb R$ and the intervals $(a, \infty)$, where $a \in \Bbb R$.

I'm confused about how should I think about the fact that $f$ is continuous with regards to the topology $\tau$, this means that for every $x \in X$, we have that $\forall V_{f(x)} \subset \Bbb R$ there is $U_x \subset X$ such that $f(U_x) \subset V_{f(x)}$.
Any hints on how to approach this?
 A: Suppose that $f$ is continuous with regards to the topology $\tau$ and take $a \in X$ and $\epsilon \gt 0$. $(f(a) - \epsilon, \infty)$ belongs to $\tau$. Therefore it exists an open subset $U$ (for the topology on $X$) with $x \in U$ such that for all $y \in U$, we have $f(y) \in (f(a) - \epsilon, \infty)$. This is exactly proving that $f$ is downward partly continuous at $a$.
Note: to ease things, it would be good to name the topology that endows $X$.
A: $\Longleftarrow$ f is continuous in topology $\tau$. Then $f^{-1}(y-\epsilon,\infty)$ is open for all $y,\epsilon \in \mathbb{R}$. Which means if we set $y=f(a)$ then we found an open set $U_a=\{x\in X : f(x)>f(a)-\epsilon\}$ for all $a$ and $\epsilon$.
$\Longrightarrow$ Let f be downward partly continuous. Then for any interval of the form $(y,\infty)$ (where $y \in \mathbb{R}$) we have $f^{-1}(y,\infty)=\{x\in X : f(x)>y\}$. But we know from our assumption that for all $a\in X$ and for all $\epsilon \in \mathbb{R}$ there is an open set $U_a=\{x\in X : f(x)>f(a)-\epsilon\}$. So you can choose and $f(a)>y$ and $\epsilon=f(a)-y$. Which means you are done. Note that if $f(a)>y$ for all $a\in X$ then the set $f^{-1}(y,\infty)$ is equal to $X$.
