Is the set of extended real-valued numbers open or closed If I assume that my topology is defined on the extended real-valued numbers, then $\mathbb{R}\cup\left\{-\infty,+\infty\right\}=\left[-\infty,+\infty\right]$, acting as my entire space, is both open and closed.
Consider now the set $\left(\alpha,+\infty\right]$. Am I right to say that because its complement $\left[-\infty,\alpha\right]$ is certainly not open (hence closed), that $\left(\alpha,+\infty\right]$ is open? Or are both $\left[-\infty,\alpha\right]$ and $\left(\alpha,+\infty\right]$ neither open nor closed? Or, are they both open and closed?
My question might appear basic, but I am a bit puzzled here...
 A: $(a,+\infty]$ is open because it contains an open neighbourhood of each of its elements. In fact, it is by definition of the topology an open neighbourhood of $+\infty$.
Hence $[-\infty,a]$ is closed. 
A: It depends on how the topology on the extended real numbers is defined. I think the standard definition of the two-point compactification is to define as a base the open intervals and the rays $(\alpha,\infty]$, $[-\infty, \beta)$, for all $\alpha,\beta\in\mathbb{R}$. In this case, the answer to your question is trivial.
Certainly a priori the set $\mathbb{R}\cup\{-\infty,\infty\}$ is not a topological space, since its topology has not been defined. To turn it into a topological space, you need to define a topology. This is a process called compactification; read the Wikipedia article for more details and examples.
A: Not all sets are either open or closed. you can' say because a set isn't open that it is closed. Usually  in topology one starts by defining the open sets, then defines closed sets as the complements of open sets.
