I am having some trouble solving the following question in Willard's General Topology (p.36).
At each point $x$ of the real line other than the origin, the basic neighbourhoods of $x$ will be the usual open intervals centred at $x$. Basic neighbourhoods of the origin will be the sets: $(-\epsilon, \epsilon) \cup (-\infty, -n) \cup (n, \infty)$, for all possible choices $\epsilon > 0$ and $n \in \mathbb{N}$.
Q1. Verify that this gives a topology on the line.
Q2. Describe the closure operation in the resulting space.
For Q1, to show that the basic neighbourhoods give a topology on $\mathbb{R}$, I have to show that the following properties hold (from Theorem 4.5 in the textbook):
Edit: as pointed out in the comments, my description of the following theorem was inaccurate before. It has been updated - thank you to Brian M. Scott for pointing this out.
a) If $V \in \mathfrak{B}_x$ then $x \in V$.
b) If $V_1, V_2 \in \mathfrak{B}_x$ then $\exists V_3 \in \mathfrak{B}_x : V_3 \subseteq V_1 \cap V_2$.
c) If $V \in \mathfrak{B}_x$ then $\exists V_0 \in \mathfrak{B}_x$ such that $\forall y \in V_0, \exists W \in \mathfrak{B}_y : W \subseteq V$.
This is not hard to show now that I have the theorem copied correctly. However, I am still having problems with Q2.
How does one go about comprehensively describing the closure operation? Based on the above, the open sets are:
- The bounded, open sets $U \subseteq \mathbb{R}$ under the standard topology, where $0 \notin U$.
- The open sets $U \subseteq \mathbb{R}$ under the standard topology, for which $0 \in U$ and $(-\infty, a) \cup (b,\infty) \subseteq U$ for some $a<0, b>0$.
Obviously, the closed sets are the complements of open sets. But I am having trouble figuring out a concise way to describe the closure operation. Any further insight is appreciated.