Open Sets in Topological Spaces

Openness Is a Property of a Topology

In Discrete and Indiscrete Topologies, the same underlying set carried two very different collections of open sets. In a general topological space, which subsets are open is determined by the chosen topology, not by the set alone. The topology axioms ensure that the collection behaves consistently under unions and finite intersections; they do not require that every subset be open or that every intersection of open sets be open.

Let $(X,\mathcal{T})$ be a topological space. A set $U\subseteq X$ is called open when $U\in\mathcal{T}$. A pointwise way to express this is that each point of $U$ has an open neighborhood contained in $U$. Here an open neighborhood of $x$ means an open set containing $x$.

Proposition: A subset $U\subseteq X$ is open if and only if, for every $x\in U$, there is an open set $V_x\in\mathcal{T}$ such that $x\in V_x\subseteq U$.

Proof. If $U$ is open, then for every $x\in U$ we can take $V_x=U$. Conversely, suppose such a set $V_x$ exists for each $x\in U$. Then $$ U=\bigcup_{x\in U}V_x. $$ Indeed, each $V_x$ is contained in $U$, so the union is contained in $U$; and each $x\in U$ belongs to its corresponding $V_x$, so $U$ is contained in the union. The topology is closed under arbitrary unions, so $U$ is open. If $U=\varnothing$, it is open by the topology axioms, and the union over no points is $\varnothing$. $\square$

This criterion makes openness local: to establish that $U$ is open, it is enough to find a suitable open neighborhood around each of its points. The neighborhoods need not be the same for different points. In a metric space, open balls often supply these neighborhoods, but the criterion itself requires no metric.

Unions, Intersections, and a Common Pitfall

The topology axioms guarantee that an arbitrary union of open sets is open and that a finite intersection of open sets is open. In particular, the intersection of two open sets is open, and repeated use gives the same conclusion for any finite number of sets. The word finite matters: an infinite intersection of open sets need not be open.

Worked Example: An Infinite Intersection That Is Not Open

Use the usual topology on $\mathbb{R}$, in which open intervals form a neighborhood around each point of an open set. For each positive integer $n$, let $$ U_n=\left(-\frac{1}{n},\frac{1}{n}\right). $$ Every $U_n$ is open. A real number $x$ belongs to every $U_n$ exactly when $|x|<1/n$ for every positive integer $n$. This holds for $x=0$. If $x\ne 0$, choose a positive integer $n>1/|x|$; then $1/n<|x|$, so $x\notin U_n$. Therefore $$ \bigcap_{n=1}^{\infty}U_n=\{0\}. $$ The singleton $\{0\}$ is not open in the usual topology: every open interval containing $0$ also contains points other than $0$. Thus an infinite intersection of open sets can fail to be open, even though each set in the intersection is open.

There is no conflict with the finite-intersection axiom. For any fixed $N$, the finite intersection $U_1\cap\cdots\cap U_N$ is $U_N$, because the intervals shrink as $n$ increases. It is open. The singleton appears only after intersecting infinitely many sets.

Worked Example: Checking Openness Point by Point

Let $X=\{a,b,c\}$, and take $$ \mathcal{T}=\{\varnothing,\{a\},\{a,b\},X\}. $$ For $U=\{a,b\}$, the point $a$ has the open neighborhood $\{a\}\subseteq U$, and $b$ has the open neighborhood $\{a,b\}\subseteq U$. The pointwise criterion therefore confirms that $U$ is open. The set $\{b\}$, however, is not open. Its only point $b$ has no open neighborhood contained in $\{b\}$: the open sets containing $b$ are $\{a,b\}$ and $X$, and neither is contained in $\{b\}$. This illustrates why a point belonging to a set does not, by itself, make that set open.

Open Sets in a Subspace

A subset can be considered as a space in its own right by using the subspace topology. If $A\subseteq X$, this topology records which parts of $A$ are inherited from open sets of $X$. In general, an open set in the subspace need not be open in the whole space.

Definition: If $(X,\mathcal{T})$ is a topological space and $A\subseteq X$, the subspace topology on $A$ is $$ \mathcal{T}_A=\{A\cap O:O\in\mathcal{T}\}. $$ A set open in $(A,\mathcal{T}_A)$ is called open in the subspace $A$.

Theorem: A set $U\subseteq A$ is open in the subspace $A$ if and only if there is an open set $O$ in $X$ such that $U=A\cap O$.

Proof. By definition, $U$ is open in $A$ exactly when $U\in\mathcal{T}_A$. Membership in $\mathcal{T}_A$ means that $U=A\cap O$ for some $O\in\mathcal{T}$, which is an open set in $X$. This proves both directions. $\square$

Worked Example: A Subspace Can Have More Open Sets Than It Inherits as Open Sets

Let $X=\{a,b,c,d\}$ have the topology consisting of unions of the two blocks $\{a,b\}$ and $\{c,d\}$. In particular, $\{a,b\}$ and $\{c,d\}$ are open in $X$. Set $A=\{a,c\}$. The intersections of $A$ with the open sets of $X$ are $$ \varnothing,\quad \{a\},\quad \{c\},\quad A. $$ Consequently, $\{a\}$ is open in the subspace $A$, since $\{a\}=A\cap\{a,b\}$. But $\{a\}$ is not open in $X$: every open set of $X$ is a union of the two blocks, and none of those unions equals $\{a\}$. Openness must therefore always be understood relative to the space being considered.

One useful special case follows directly from the theorem. If $A$ itself is open in $X$, then every open subset $U$ of the subspace $A$ is open in $X$. In fact, write $U=A\cap O$ for an open $O\subseteq X$. Both $A$ and $O$ are open in $X$, so their finite intersection $U$ is open in $X$. Without the assumption that $A$ is open, the conclusion can fail, as the example shows.

Bases: Describing Open Sets by Smaller Pieces

A topology may contain many open sets, but it can sometimes be described efficiently using a smaller collection of open sets. The idea is to use these smaller sets as building blocks: every open set should be a union of them. Two conditions ensure that these building blocks fit together consistently.

Definition: A collection $\mathcal{B}$ of subsets of $X$ is a basis for a topology on $X$ if (1) every $x\in X$ belongs to at least one member of $\mathcal{B}$, and (2) whenever $x\in B_1\cap B_2$ for $B_1,B_2\in\mathcal{B}$, there is $B_3\in\mathcal{B}$ such that $x\in B_3\subseteq B_1\cap B_2$.

The first condition ensures that the building blocks cover the space. The second says that where two blocks overlap, points in the overlap still have a basis block around them that stays within the overlap. A basis set need not itself be a singleton, and it need not be the only basis set containing a given point.

Theorem: If $\mathcal{B}$ satisfies the two basis conditions, then the collection $$ \mathcal{T}_{\mathcal{B}}=\left\{\bigcup_{B\in\mathcal{C}}B:\mathcal{C}\subseteq\mathcal{B}\right\} $$ is a topology on $X$. Its open sets are exactly the unions of members of $\mathcal{B}$.

Proof. The empty set is the union of the empty subcollection, so $\varnothing\in\mathcal{T}_{\mathcal{B}}$. By the covering condition, $X$ is the union of all members of $\mathcal{B}$, so $X\in\mathcal{T}_{\mathcal{B}}$. An arbitrary union of sets that are themselves unions of basis members is again a union of basis members. Thus $\mathcal{T}_{\mathcal{B}}$ is closed under arbitrary unions.

It remains to check finite intersections. Let $U=\bigcup_{B\in\mathcal{C}}B$ and $V=\bigcup_{D\in\mathcal{D}}D$ be in $\mathcal{T}_{\mathcal{B}}$. If $x\in U\cap V$, then there are $B\in\mathcal{C}$ and $D\in\mathcal{D}$ with $x\in B\cap D$. The intersection condition gives a $G\in\mathcal{B}$ such that $x\in G\subseteq B\cap D\subseteq U\cap V$. Hence every point of $U\cap V$ belongs to a basis member contained in $U\cap V$, and $$ U\cap V=\bigcup\{G\in\mathcal{B}:G\subseteq U\cap V\}. $$ The right-hand side is a union of basis members, so $U\cap V\in\mathcal{T}_{\mathcal{B}}$. This proves closure under intersections of two sets. Repeating the argument proves closure under any positive finite number of intersections; the intersection of no sets is $X$, already included. Therefore $\mathcal{T}_{\mathcal{B}}$ is a topology. Its definition says precisely that its open sets are unions of members of $\mathcal{B}$. $\square$

Worked Example: Constructing a Topology from a Basis

On $X=\{a,b,c\}$, consider $$ \mathcal{B}=\{\{a,b\},\{b,c\},\{b\}\}. $$ Every point belongs to a member of $\mathcal{B}$: $a$ belongs to $\{a,b\}$, $b$ belongs to all three members, and $c$ belongs to $\{b,c\}$. The only overlap between distinct two-point members is $\{a,b\}\cap\{b,c\}=\{b\}$, which is itself in $\mathcal{B}$. Intersections involving $\{b\}$ are either $\{b\}$ or empty, and the basis condition only asks for a smaller basis member at points that actually lie in the intersection. Thus the conditions hold.

The unions of basis members are $$ \varnothing,\quad \{b\},\quad \{a,b\},\quad \{b,c\},\quad X. $$ For example, $\{a,b\}\cup\{b,c\}=X$, while neither $\{a\}$ nor $\{c\}$ is a union of members of $\mathcal{B}$. The theorem guarantees that the displayed collection is a topology and that its open sets are exactly these unions.

For an already specified topology $\mathcal{T}$, the same idea gives a practical test: a collection $\mathcal{B}\subseteq\mathcal{T}$ is a basis for $\mathcal{T}$ precisely when every open set in $\mathcal{T}$ is a union of members of $\mathcal{B}$. The basis conditions encode how such a collection can be used to generate a topology; the union description explains how it represents open sets.

Choosing the Right View of Openness

The pointwise criterion is useful when proving that one particular set is open: find an open neighborhood inside the set for each of its points. The subspace criterion is useful when the space has been restricted: express an open set as the intersection of the subspace with an ambient open set. A basis is useful when many open sets share a small collection of building blocks.

Keep the scope of each statement clear. Arbitrary unions of open sets are open, but only finite intersections are guaranteed to be open. A set open in a subspace need not be open in the ambient space. And having a family of subsets that covers $X$ is not enough by itself to make it a basis: the overlap condition must also hold. These distinctions are central to using open sets reliably in general topological spaces.

Check Your Understanding

Use the definitions and results in this tutorial to answer the following questions.

How can the pointwise neighborhood criterion be used to prove that a subset is open?
Why does the intersection of the intervals $(-1/n,1/n)$, over all positive integers $n$, fail to be open in the usual topology on $\mathbb{R}$?
What form must every open set in a subspace $A\subseteq X$ have?
In the finite example with $A=\{a,c\}$, why is $\{a\}$ open in $A$ but not open in $X$?
State the two conditions a collection must satisfy to be a basis.
Why does the intersection condition for a basis help prove that the generated collection is closed under intersections of two open sets?