Topology Closed Neighborhood Proof

(8) N is a closed neighborhood of p in R_U iff N=R

Where R represents the real numbers and

R_U is the upper topological space for the real numbers (R, T_U)

T_U = {A is a subset of R | a is an element of A implies (a - e, +inf) is a subset of A for some e > 0} (upper topology)

I proved "=>"

Need help showing "<="

ie: If N=R, then N is a closed neighborhood of p in R_U

Re: Topology Closed Neighborhood Proof

The topology of $\displaystyle \mathbb{R}_U$ is the set $\displaystyle \{ \emptyset \} \cup \{ \mathbb{R} \} \cup \{ (r, \infty) | r \in \mathbb{R} \}$.

The direction you did was the "hard" direction. I think you forgot that the whole space is always a closed set, since it's the complement of the empty set, which is always open.

Let $\displaystyle p \in \mathbb{R}$. Then $\displaystyle p \in (p-1, \infty)$, which is open. Let $\displaystyle N = \mathbb{R}$. Will show $\displaystyle N$ is a closed neighborhood of $\displaystyle p$.

Have that $\displaystyle N = \{\emptyset\}^c$ is closed. Since $\displaystyle p \in (p-1, \infty) \subset N$, it follows that $\displaystyle N$ is a closed neighborhood of $\displaystyle p$.