# How do I prove that $(c,\infty]$ is open in $\overline{\mathbb{R}}$?

$\overline{\mathbb{R}}$ is a topological space, but not a metric space, so I'm not sure if it is true.

Let $U$ be an open set in $\overline{\mathbb{R}}$ such that $\infty\in U$.

Then, how do I that prove there exists $c\in\mathbb{R}$ such that $(c,\infty] \subset U$ and such that $(c,\infty]$ is an open set?

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What is the definition of the topology on $\overline{\mathbb R}$? (There is a standard topology, but knowing your precise definition makes it easier to answer your question, and your close examination of the definition may also lead you to answer your own question.) –  Jonas Meyer Dec 7 '12 at 4:13
@Jonas To be honest, i'm not sure which topology it should be for $\overline{\mathbb{R}}$. Would you tell me what is the standard topology on $\overline{\mathbb{R}}$? Neighborhoods? –  Katlus Dec 7 '12 at 4:23
Standard topology is all the usual open sets, and then a set containing $\infty$ is open if and only if its complement is a compact set in the usual topology of $\mathbb{R}$. –  rondo9 Dec 7 '12 at 4:32
By $\overline{\mathbb{R}}$ do you mean $\mathbb{R} \cup \{ \infty \}$ or $\mathbb{R} \cup \{ - \infty , + \infty \}$, or something else? –  Arthur Fischer Dec 7 '12 at 4:50
@Arthur I meant $\mathbb{R} \cup \{\infty,-\infty\}$. I think i got the answer. "$U$ is open iff $\forall x\in U, [x\in \mathbb{R} \Rightarrow \exists \epsilon>0 B(x,\epsilon)\subset U]\bigvee [x=\infty \Rightarrow \exists c\in \mathbb{R} (c,\infty]\subset U] \bigvee [x=-\infty \Rightarrow \exists c\in \mathbb{R} [-\infty,c)\subset U]$". Am i correct? –  Katlus Dec 7 '12 at 6:11