I am trying to prove that if $\mathcal U$ is a non-convergent ultrafilter on $[0,\infty)$ with its usual topology, then $\mathcal U$ must contain intervals of the form $[a,\infty)$ for every $a \ge 0$.
I am trying to argue by contradiction: Suppose that for some $a \ge 0$, $\mathcal U$ does not contain $[a,\infty)$. Since $\mathcal U$ is an ultrafilter, it must therefore contain $[0,a)$. I want to argue somehow that $\mathcal U \to a$, contradicting that $\mathcal U$ is non-convergent. Thus, I need to show that $\mathcal U$ contains every interval of the form $(c,d)$ with $a \in (c,d)$. Given such a $(c,d)$, I know that $[0,d) \in \mathcal U$ since $[0,a) \subseteq [0,d)$. Somehow I need to show that $\mathcal U$ contains $(c,\infty)$ or something comparable to it, in which case $\mathcal U$ will contain $(c,\infty) \cap [0,d) = (c,d)$. I have tried arguing by contradiction again, but I seem to be getting nowhere. Any tips?