Topology Closed Neighborhood Proof

**Kiefer** · September 13th 2012, 04:59 PM

(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

**johnsomeone** · September 13th 2012, 05:43 PM

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

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

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

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