# Non-convergent ultrafilter on $[0,\infty)$

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?

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I think that you don't want to use some $a$. You want to take the set $\{x\in[0,\infty)\mid [x,\infty)\notin\cal U\}$, and argue that since it isn't empty, it must have an infimum, and take that to be your $a$. I mean, note that if $[1,\infty)\notin\cal U$ then certainly $[2,\infty)\notin\cal U$. But it is impossible that $\cal U$ converges to both points, since the space is Hausdorff. – Asaf Karagila Jun 5 '13 at 13:19

I think your start is OK: we have some $a > 0$ such that $[0,a) \in \mathcal{U}$, and so also $[0,a] \in \mathcal{U}$, as this is an enlargement.
Now $\mathcal{U}' = \{ A \cap [0,a]: A \in \mathcal{U} \}$ is an ultrafilter on the compact set $[0,a]$, so it converges to some unique $p \in [0,a]$. But then $\mathcal{U}$ also converges to $p$ and this contradicts that $\mathcal{U}$ is not convergent.
You may want to take the infimum of all $a$'s so that $[0,a)$ is included in $\mathcal U$.