Topological space

Revision as of 15:38, 21 June 2008 by Boy Soprano II (talk | contribs) (links to similar notions)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

A topological space (or simply a topology) is an pair $(X,\mathcal{T}$), where $X$ is a set and $\mathcal{T}$ is a subset of the power set of $X$ satisfying the following relations:

By abuse of language, the topology $(X,\mathcal{T})$ is called simply "the topology $X$" or "the topology $\mathcal{T}$", depending on context.

The elements of $\mathcal{T}$ are usually called open sets. The subsets of $X$ whose complements are elements of $\mathcal{T}$ are then called closed sets.

It follows from the principles of set theory that the intersection of any family of closed sets is closed, and that the intersection of any family of open sets is open.

When checking whether a set $\mathcal{T}$ is a topology, by induction it suffices to show that the intersection of any two elements of $\mathcal{T}$ is also an element of $\mathcal{T}$.


Trivially, the sets $\{ \varnothing, X\}$, and $\mathfrak{P}(X)$ (the power set of $X$) are both topologies. They are the least and greatest topologies on $X$, respectively.

If $(\mathcal{T}_i)_{i\in I}$ is a family of topologies on $X$, then so is their intersection, $\bigcap_{i\in I} \mathcal{T}_i$.

Let $\mathcal{Q}$ be a collection of subsets of $X$. Since $\mathcal{P}(X)$ is a topology on $X$ of which $\mathcal{Q}$ is a subset, the set of topologies on $X$ that contain $\mathcal{Q}$ is not empty. It then follows from the previous paragraph that there exists a least topology on $X$ containing $\mathcal{Q}$. This is the topology generated by $\mathcal{Q}$.

Theorem. Let $\mathcal{Q}$ be a set with the property: for any $A,B$ in $\mathcal{Q}$ and any $x\in A\cap B$, there exists an element $C$ of $\mathcal{Q}$ such that \[x \in C \subset A \cap B .\] Then the topology generated by $\mathcal{Q}$ is the set containing $X$, and the subsets $S$ of $\bigcup_{A\in \mathcal{Q}} A$ with the following property: for every element $x$ of $S$, there exists an element $A$ of $\mathcal{Q}$ such that $x\in A$ and $A\subset S$.

Proof. Call the set described in the theorem $\mathcal{T}$. From the properties of topologies it is evident that every topology containing $\mathcal{Q}$ must also contain $\mathcal{T}$. Hence it suffices to show that $\mathcal{T}$ is in fact a topology.

By construction, $X$ is an element of $\mathcal{T}$; since $\varnothing$ has no elements, it vacuously satisfies the theorem's condition and hence is also an element of $\mathcal{T}$.

Let $(T_i)_{i\in I}$ be a family of elements of $\mathcal{T}$. If one of them is $X$, then their union is $X$, which is an element of $\mathcal{T}$. Otherwise, we note that every element $x$ of $\bigcup_{i\in I} T_i$ is an element of $T_i$; hence there exists some $A\in \mathcal{Q}$ for which \[x\in A \subset T_i \subset \bigcup_{i\in I} T_i .\]

Finally, suppose $T_1$ and $T_2$ are two elements of $\mathcal{T}$. If one, say $T_1$, is equal to $X$, then $T_1 \cap T_2 = T_2 \in \mathcal{T}$, and we are done. Otherwise, for every element $x$ of $T_1 \cap T_2$, there exist sets $A\subset T_1$ and $B \subset T_2$ for which $x \in A\cap B$. By hypothesis, there exists a set $C$ in $\mathcal{Q}$ such that \[x \in C \subset A \cap B \subset T_1 \cap T_2 .\] This proves that $T_1 \cap T_2 \in \mathcal{Q}$. Hence $\mathcal{T}$ is a topology, as desired. $\blacksquare$

In a metric space $X$, we can define an open set to be a set $S$ such that for every element $x \in S$ there exists an open ball centered at $x$ that is contained in $S$. This is the topology generated by the neighborhoods of $X$, and it is called the metric topology of $X$. The closed sets of $X$ are then those sets $S$ such that every limit point of $S$ is an element of $S$.

In $\mathbb{Z}$, the topology generated by the infinite arithmetic sequences yields an interesting proof of the infinitude of primes. Specifically, consider the topology generated by the residue classes modulo $m$, as $m$ ranges through all nonnegative integers. Then for every integer $n$, the set $n \mathbb{Z}$ is closed and open. If there are finitely many primes $p$, it then follows that the union of all sets $p\mathbb{Z}$ is closed. Since every integer except $\pm 1$ is divisible by some prime, it follows that the set $\{-1,1\}$ is open, a contradiction.


Let $X$ be a set with a topology $\mathcal{T}$, and let $S$ be a subset of $X$. The topology $\mathcal{T}$ induces a topology on $S$, namely the set of sets of the form $S \cap T$, for $T \in \mathcal{T}$. This topology is called the topological subspace $S$.

Note that open and closed sets in the subspace $S$ are not necessarily open or closed in the space $X$.

See also

Invalid username
Login to AoPS