Does the Tietze extension theorem hold for maps into R if on R we put the semi open interval topology? I guess the above is false, so I'm trying to find a counter example. The only thing I found is the whole space should not be compact. If not, the statement is true directly from the Tietze extension.
Any help will be appreciated.
 A: No, this does not hold, as $\mathbb{R}$ in the lower limit topology $\mathcal{T}_l$, generated by all sets of the form $(a,b]$ (or homeomorphically by sets $[a,b)$ if one prefers) (a. k.a. the Sorgenfrey line) is zero-dimensional and thus totally disconnected. 
This implies that any continuous function $f : X \to \mathbb{R}$ is constant whenever $X$ is connected. So for connected normal spaces we don't have Urysohn functions to separate the closed sets.
A: I guess you mean the topology generated by the half-open intervals $[a,b)$. It's called the lower limit topology: https://en.wikipedia.org/wiki/Lower_limit_topology
You are asking if the Tietze extension theorem still holds on a space $X$ (I assume normal) if the topology on $R$ is not the standard topology (as it is in the usual Tietze extension theorem) but is instead the half-open interval topology. 
I think the semi-open Tietze extension property will imply a semi-open Urysohn lemma property in the usual way: http://at.yorku.ca/cgi-bin/bbqa?forum=ask_a_topologist_2001&task=show_msg&msg=0263.0001
But R with the lower limit topology is completely disconnected. So any continuous function from a connected space $X$ to $R$ will be constant. So, if $X$ is connected, a semi-open Urysohn lemma cannot hold, hence a semi-open Tietze theorem cannot hold. 
