Open Sets

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Open Sets and Topology

Algebraic structures need not carry well-defined notions such as distance, which we usually associate with the word space. On the other hand, the spaces mathematicians care about can differ markedly from these familiar examples; for instance, many spaces cannot be assigned a distance at all. In this post we examine one way to define a space in full generality.

Definition 1 A topology on a set $X$ is a subset $\mathcal{T}$ of $\mathcal{P}(X)$ satisfying the following conditions.

$\emptyset,X\in\mathcal{T}$.
The arbitrary union of elements of $\mathcal{T}$ belongs to $\mathcal{T}$.
The finite intersection of elements of $\mathcal{T}$ belongs to $\mathcal{T}$.

A set $X$ equipped with such a topology is called a topological space; its elements are called points, and the elements of $\mathcal{T}$ are called open sets.

Adopting the convention from §Union and Intersection, one can show that the two sets

\[\bigcup_{U\in\emptyset}U=\emptyset,\qquad\bigcap_{U\in\emptyset}U=X\]

must belong to $\mathcal{T}$ by the second and third conditions alone, but it is customary to state the first condition explicitly to avoid confusion.

Example 2 For any set $X$, the smallest topology on $X$ is $\{\emptyset,X\}$. One readily checks that this family satisfies conditions 1–3, so it yields a topological space. It is called the trivial topology.

Conversely, the largest topology on $X$ is $\mathcal{P}(X)$. This also forms a topological space, called the discrete topology.

In general, when there is no risk of confusion about the topology on $X$, we simply write the topological space $X$.

Definition 3 Let two topologies $\mathcal{T}_1$ and $\mathcal{T}_2$ on a set be comparable under $\subset$. If $\mathcal{T}_1\subset\mathcal{T}_2$, then $\mathcal{T}_2$ is called a stronger topology than $\mathcal{T}_1$, and $\mathcal{T}_1$ is called a weaker topology than $\mathcal{T}_2$.

For example, any topology $\mathcal{T}$ on a set $X$ is always weaker than the discrete topology on $X$ and always stronger than the trivial topology on $X$.

The best tool for introducing a notion of nearness on $X$ is distance, but since only very few spaces admit a distance, a more general notion is needed. One can recover a kind of nearness from $\mathcal{T}$ by examining how many open sets contain two given points in common; this yields a theory similar to the metric one.

First consider the trivial topology. If $X=\{x,y\}$ and $X$ carries the trivial topology, then the two points $x,y$ are perfectly indistinguishable from the topological point of view. This is because the topological structure cannot be used¹ to tell $x$ and $y$ apart. When two points $x,y$ of a topological space $X$ are contained in exactly the same open sets, we say that they are topologically indistinguishable.

On the other hand, if the same set $X$ is given the discrete topology, then the points $x$ and $y$ become distinguishable: the open set $\{x\}$ contains $x$ but not $y$. In this sense, all points are glued together in the trivial topology, whereas they are isolated from one another in the discrete topology.

Since an open set containing the point $x$ will be needed frequently, it is convenient to introduce the following terminology.

Definition 4 Let $X$ be a topological space and let $A$ be an arbitrary subset of $X$. Any open set containing $A$ is called an open neighborhood of $A$. A subset $S$ of $X$ is called a neighborhood of $A$ if there exists an open neighborhood $U$ of $A$ such that $U\subseteq S$.

In particular, when $A$ is a singleton $\{x\}$, the open neighborhoods and neighborhoods of $A$ are also called the open neighborhoods and neighborhoods of $x$, respectively.

In many cases it suffices to study open neighborhoods alone, and some texts such as [Mun] simply call them neighborhoods. However, distinguishing the two notions is essential for a clean formulation of several results in topology. Since this extra generality poses no real difficulty, we shall keep the distinction explicit.

Proposition 5 A subset $U$ of a topological space $X$ is open if and only if $U$ is a neighborhood of every $x\in U$.

Proof

By Definition 4, the latter condition is equivalent to:

For every $x\in U$, there exists an open neighborhood $V$ of $x$ such that $V\subseteq U$.

If $U$ is open, then $U$ itself is an open neighborhood of $x$ contained in $U$, so the forward direction is immediate. It therefore suffices to prove the converse. Suppose that for every $x\in U$ there exists an open neighborhood of $x$ contained in $U$; to indicate the dependence on $x$ we denote it by $V(x)$. Thus each $V(x)$ is an open set satisfying $\{x\}\subseteq V(x)\subseteq U$. That $U$ is open now follows from the chain

\[U=\bigcup_{x\in U}\{x\}\subseteq\bigcup_{x\in U} V(x)\subseteq\bigcup_{x\in U} U=U.\]

Neighborhood filter

Let $X$ be a topological space and $x\in X$. Define $\mathcal{N}(x)$ to be the collection of all neighborhoods of $x$.

Proposition 6 Let $X$ be a topological space. Then $\mathcal{N}(x)$ satisfies the following properties.

Any set containing an element of $\mathcal{N}(x)$ belongs to $\mathcal{N}(x)$.
The finite intersection of any elements of $\mathcal{N}(x)$ again belongs to $\mathcal{N}(x)$.
Every element of $\mathcal{N}(x)$ contains $x$.
For every element $V\in\mathcal{N}(x)$, there exists a suitable $W\in\mathcal{N}(x)$ such that for any $y\in W$ we have $V\in\mathcal{N}(y)$.

Conversely, suppose that for each element $x\in X$ of an arbitrary set $X$, a collection $\mathcal{N}(x)\subseteq\mathcal{P}(X)$ satisfying the above conditions is given.² Then there exists a unique topology $\mathcal{T}$ on $X$ such that these $\mathcal{N}(x)$ are precisely the collections of all neighborhoods of $x$.

Proof

The first and third properties are immediate from the definitions.

For the second property, let $V_1,\ldots, V_n\in\mathcal{N}(x)$ be given. Then there exist open neighborhoods $U_1,\ldots, U_n$ of $x$ with $x\in V_i\subseteq U_i$. By the third condition of Definition 1, their finite intersection $U_1\cap\cdots\cap U_n$ is also open, and therefore

\[x\in \bigcap_{i=1}^n U_i\subseteq\bigcap_{i=1}^n V_i,\]

so $\bigcap V_i$ is a neighborhood of $x$.

For the fourth property, we may simply take $W$ to be an open neighborhood of $x$ contained in $V$.

Now we must show the converse. Suppose collections $\mathcal{N}(x)$ satisfying the above conditions are given. By Proposition 5, the elements of $\mathcal{T}$ must satisfy the condition

For every $x\in U$, $U\in\mathcal{N}(x)$.

Hence $\mathcal{T}$ is uniquely determined.

To show existence, let $\mathcal{T}$ be the collection of all sets satisfying the above condition; we verify that this family satisfies the conditions of Definition 1. First, it is clear that $\emptyset$ and $X$ satisfy the condition. Let $(U_i)_{i\in I}$ be elements of $\mathcal{T}$, let $U=\bigcup U_i$, and pick an arbitrary $x\in U$. Then $x\in U_i$ for some $i$; since $U_i\in\mathcal{N}(x)$ and $U_i\subseteq U$, the first condition yields $U\in\mathcal{N}(x)$, and therefore $U\in\mathcal{T}$. Similarly, one shows that the intersection of finitely many elements of $\mathcal{T}$ belongs to $\mathcal{T}$.

To complete the proof of existence, we must show that for the topological space $(X,\mathcal{T})$ thus constructed, $\mathcal{N}(x)$ actually coincides with the collection of all neighborhoods of each point $x$. By the definition of $\mathcal{T}$, any open neighborhood $U$ of $x$ lies in $\mathcal{N}(x)$, and hence any neighborhood of $x$ lies in $\mathcal{N}(x)$ by the first condition.
Conversely, let $V$ be an arbitrary set in $\mathcal{N}(x)$. To see that $V$ is a neighborhood of $x$ in $\mathcal{T}$, we must find an open neighborhood $U\subseteq V$ of $x$. Define $U$ as the set of all $y\in X$ such that $V\in\mathcal{N}(y)$.

Since $V\in\mathcal{N}(x)$, we clearly have $x\in U$.
Pick an arbitrary $y\in U$. By definition of $U$ we have $V\in\mathcal{N}(y)$, and the third condition for $\mathcal{N}(y)$ gives $y\in V$. Thus $x\in U\subseteq V$.
Finally we must show that $U\in\mathcal{T}$, i.e. that $U\in\mathcal{N}(y)$ for every $y\in U$. Since $V\in\mathcal{N}(y)$, the fourth condition yields a suitable $W\in\mathcal{N}(y)$ such that $V\in\mathcal{N}(z)$ for every $z\in W$. Hence $z\in U$, so $W\subseteq U$, and therefore $U\in\mathcal{N}(y)$.

The first and second properties show that $\mathcal{N}(x)$ is a filter on the ordered set $(\mathcal{P}(X),\subseteq)$, and the third property shows that this filter does not contain the empty set $\emptyset$. A subset of $\mathcal{P}(X)$ satisfying these three conditions is called a filter on the topological space $X$. In particular, the filters $\mathcal{N}(x)$ are called neighborhood filters.

References

[Bou] N. Bourbaki, General Topology. Elements of mathematics. Springer, 1995.

That is, using the open sets that are elements of $\mathcal{T}$ ↩
The second condition is understood to imply $\bigcap_{V\in\emptyset} V=X\in\mathcal{N}(x)$. ↩

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Open Sets and Topology

Neighborhood filter

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