# Definition of Topology on $\mathbb{R}$ via Neighbourhoods

A topology can be defined via different axioms. One of them see here, in terms of neighbourhoods, states a topology is a set $X$ endowed with neighbourhood function $N:X\to\mathcal{F}(X)$, where $\mathcal{F}(X)$ denotes the set of all filters on $X$, satisfying the following two conditions:

1. $U\in N(x)\implies x\in U$
2. $U\in N(x)\implies \exists V\in N(x),\ \forall y\in V: U \in N(y)$

Furthermore, an open set is a set if it is a neighbourhood of each of its points.

I am not quite understand how this definition implies the open set definition of topology, say arbitary union of open sets is open set.

Also, I cannot find out any inconsistency if I choose all closed intervals as open sets on $\mathbb{R}$.

• Arbitrary unions of open sets are open. Only intersections of finitely many open sets are guaranteed to be open. Sep 18 '15 at 14:51

I am not quite understand how this definition implies the open set definition of topology, say arbitrary union of open sets is open set.

The set of neighborhoods of $x$ which are neighborhoods of their elements is closed under arbitrary unions since the neighborhoods at $x$ are an upward closed set. Those sets form a base for the topology.

I cannot find out any inconsistency if I choose all closed intervals as open sets on ℝ

If you are using the set of closed intervals with nonequal endpoints, then the collection of closed intervals lying above an element $x\in X$ is not a filter. Clearly $[0,1/2]$ and $[1/2,1]$ would be in the neighborhoods of $1/2$, but what closed interval with nonequal endpoints is contained in both? This is why this choice does not result in a consistent topology.

On the other hand if you allow degenerate closed intervals that are points, the filter axioms are all satisfied and you have chosen valid neighborhood filters. But you have now caused all points to be open sets, and the topology is discrete.

• I understand now. Thank you. But the strange thing is that whether I can remove the second axiom if I focus only on open neighbourhood? Sep 18 '15 at 23:39

So suppose we have a filter $N(x)$ for every $x \in X$ that satisfies the two axioms.

Then define a supposed topology as all sets $O$ such that $\forall x \in O: O \in N(x)$, which is the formal way of stating a set is open iff it is a neighbourhood of each of its points.

Clearly $\emptyset$ is open (a universal quantifier over an empty domain..) and $X$ is in any filter on $X$ (filters are closed under supersets), so $X$ is also open.

If $O_i$, $i \in I$ are all open, then pick $x \in O = \cup_{i \in I} O_i$. Then $x \in O_j$ for some $j \in I$. As $O_j$ is open, $O_j \in N(x)$, but then also $O \in N(x)$, again because filters are closed under supersets. But as $x$ is arbitrary, $O$ is open as well. This takes care of unions.

For finite intersections, it suffices to show the case of two open sets $O_1,O_2$. Let $x \in O_1 \cap O_2$. Then $O_1 \in N(x)$ as $O_1$ is open and $O_2 \in N(x)$ as well, so as $N(x)$ is a filter, so closed under finite intersections, we know that $O_1 \cap O_2 \in N(x)$ as required. So $O_1 \cap O_2$ is open.

So we have a topology and the last axiom is indeed not used in the proof that this is a topology. I believe it is needed to show that $N(x)$ that we stared with then becomes exactly the neighbourhood filter of $x$ in this newly defined topology (where $N \subset X$ is a neighbourhood of $x$ iff there exists an open set $O$ such that $x \in O \subseteq N$). This could be a nice exercise, perhaps?

The remark about closed intervals is somewhat strange: you have to define a filter for each $x$, and closed intervals by themselves do not form a filter. They do form a filter base (so all closed intervals that contain $x$ and all their supersets), but then we have to allow trivial intervals $[x,x] = \{x\}$ for this to work, and then we get the dicrete topology. We could define $N(x)$ as all supersets of open intervals that contain $x$ and then we'd get the usual topology.

• I think the last axiom states every neighbourhood contains an open set. So maybe it is not necessary if, as some topologist requires, neighbourhoods are open? Sep 18 '15 at 23:42