Subspace Topology

How a Subset Inherits a Topology

A subset of a topological space may have open sets that are not open in the whole space. The key is to measure openness relative to the subset: a set is open there when it is the intersection of the subset with an open set of the ambient space. This lets us treat intervals, curves, and other subsets as topological spaces in their own right, without changing the open-set structure they inherit.

For example, the interval $[0,1]$ is not open in $\mathbb{R}$ with its usual topology. But $[0,1]$ is open as a subset of itself, and sets such as $[0,1/2)$ are open in $[0,1]$. To make these statements precise, we first construct the topology on a subset and then examine how it relates to the ambient topology.

Definition: Let $(X,\mathcal{T})$ be a topological space and $A\subseteq X$. The subspace topology on $A$ is the collection $\mathcal{T}_A=\{A\cap O:O\in\mathcal{T}\}$. The set $A$, equipped with $\mathcal{T}_A$, is called a subspace of $X$.

The open sets of $A$ are therefore traces of open sets of $X$ on $A$. The open-set characterization of subspaces established earlier in this course says equivalently that $U\subseteq A$ is open in $A$ exactly when $U=A\cap O$ for some open $O\subseteq X$. The ambient open set $O$ need not be unique.

Why the Inherited Collection Is a Topology

The definition gives a collection of subsets of $A$. It must also satisfy the topology axioms. The next result verifies this construction and is useful whenever a subset is given the topology inherited from a larger space.

Theorem: If $(X,\mathcal{T})$ is a topological space and $A\subseteq X$, then $\mathcal{T}_A=\{A\cap O:O\in\mathcal{T}\}$ is a topology on $A$.

Proof. Since $\varnothing$ and $X$ are open in $X$, we have $A\cap\varnothing=\varnothing$ and $A\cap X=A$, so $\varnothing,A\in\mathcal{T}_A$. Let $\{U_\lambda:\lambda\in\Lambda\}$ be any family of sets in $\mathcal{T}_A$. For each $\lambda$, choose an open $O_\lambda\subseteq X$ such that $U_\lambda=A\cap O_\lambda$. Then $$ \bigcup_{\lambda\in\Lambda}U_\lambda =\bigcup_{\lambda\in\Lambda}(A\cap O_\lambda) =A\cap\left(\bigcup_{\lambda\in\Lambda}O_\lambda\right). $$ The union on the right inside the parentheses is open in $X$, so this union of the $U_\lambda$ belongs to $\mathcal{T}_A$. This also covers the empty family, whose union is $\varnothing$.

Now let $U_1,\ldots,U_n\in\mathcal{T}_A$, where $n$ is a positive integer, and write $U_j=A\cap O_j$ with each $O_j$ open in $X$. Then $$ \bigcap_{j=1}^{n}U_j =\bigcap_{j=1}^{n}(A\cap O_j) =A\cap\left(\bigcap_{j=1}^{n}O_j\right). $$ A finite intersection of open sets in $X$ is open, so this finite intersection belongs to $\mathcal{T}_A$. All three topology axioms hold. $\square$

Worked Example: The Usual Topology on a Closed Interval

Give $A=[0,1]$ the subspace topology inherited from $\mathbb{R}$. The set $[0,1/2)$ is open in $A$, because $$ [0,1]\cap(-1,1/2)=[0,1/2), $$ and $(-1,1/2)$ is open in $\mathbb{R}$. The endpoint $0$ belongs to this relative open set, even though no open interval around $0$ in $\mathbb{R}$ lies inside $[0,1/2)$.

In contrast, $(0,1/2)$ is also open in $A$, since it is the intersection of $A$ with the ambient open set $(0,1/2)$. Thus relative openness does not require every point to have an ambient neighborhood contained in the set; it requires an ambient open set whose intersection with $A$ is the desired set.

Open and Closed Sets in a Subspace

The relative-open description has a parallel for closed sets. The closed-set characterization of subspaces established earlier in the course says that $F\subseteq A$ is closed in $A$ if and only if $F=A\cap C$ for some closed set $C\subseteq X$. In particular, being closed in a subspace does not generally mean being closed in the ambient space.

Worked Example: A Set Closed in a Subspace but Not in the Ambient Space

Let $A=(0,2)\subseteq\mathbb{R}$ and $F=(0,1]\subseteq A$. The set $F$ is not closed in $\mathbb{R}$, since $0$ is a limit point of $F$ but $0\notin F$. Nevertheless, $F$ is closed in $A$. Indeed, $$ F=A\cap[0,1], $$ and $[0,1]$ is closed in $\mathbb{R}$. This is exactly the closed-set characterization for the subspace $A$.

The converse distinction also matters: if a set is closed in the ambient space and contained in $A$, then it is closed in $A$, because it is its own intersection with $A$. But an ambient set that is not closed may still have a closed trace on $A$, as this example shows.

A subset $A$ itself is always both open and closed as a subset of the subspace $A$, since it is the whole space there. This does not imply that $A$ is open or closed in $X$. The words “open” and “closed” always refer to a particular space unless the ambient space is specified.

Bases and Nested Subspaces

A basis for the ambient topology gives a convenient basis for every subspace: intersect each basis element with the subset. This is the subspace-basis theorem established earlier. It often replaces a direct analysis of every open set by a simpler analysis of basic open sets.

Worked Example: The Integers Inherit the Discrete Topology

Consider $\mathbb{Z}\subseteq\mathbb{R}$, with the usual topology on $\mathbb{R}$. For each integer $n$, the interval $(n-1/2,n+1/2)$ is open in $\mathbb{R}$, and $$ \mathbb{Z}\cap(n-1/2,n+1/2)=\{n\}. $$ There is no other integer in that interval: if $m\in\mathbb{Z}$ and $m\ne n$, then $|m-n|\geq 1$, so $m$ cannot lie within distance $1/2$ of $n$. Thus each singleton $\{n\}$ is open in the subspace $\mathbb{Z}$.

Every subset of $\mathbb{Z}$ is a union of its singleton sets, so every subset is open in $\mathbb{Z}$. Therefore $\mathbb{Z}$ with its subspace topology is discrete, even though $\mathbb{R}$ is not discrete.

Subspaces can be taken in stages. If $B\subseteq A\subseteq X$, we may first give $A$ the topology inherited from $X$, and then give $B$ the topology inherited from $A$. The result is the same as giving $B$ the subspace topology directly from $X$.

Theorem: Let $B\subseteq A\subseteq X$, where $A$ has the subspace topology from $X$. The subspace topology on $B$ inherited from $A$ is the same as the subspace topology on $B$ inherited directly from $X$.

Proof. Let $O$ be open in $X$. The set $A\cap O$ is open in $A$, and its trace on $B$ is $$ B\cap(A\cap O)=B\cap O, $$ because $B\subseteq A$. Thus every set open in $B$ when inherited directly from $X$ is open when inherited through $A$.

For the reverse inclusion, let $V$ be open in $B$ when $B$ inherits its topology from $A$. By the definition of the subspace topology, $V=B\cap U$ for some open $U\subseteq A$. Since $A$ has the subspace topology from $X$, there is an open $O\subseteq X$ such that $U=A\cap O$. Therefore $$ V=B\cap(A\cap O)=B\cap O. $$ So $V$ is open in $B$ when the topology is inherited directly from $X$. The two collections of open sets are equal, as required. $\square$

This transitivity means that the topology on a subset depends only on its inclusion in the original space, not on whether we describe that inclusion in one step or through an intermediate subset. It is especially useful when a space is built from smaller pieces and those pieces are themselves subsets of other spaces.

Closure Relative to a Subspace

Closure is also sensitive to the space in which it is computed. If $E\subseteq A\subseteq X$, the closure of $E$ in $A$ consists of points of $A$ that cannot be separated from $E$ by an open neighborhood in $A$. It is related to the ambient closure by intersecting that closure with $A$.

Theorem: If $E\subseteq A\subseteq X$, then the closure of $E$ in the subspace $A$ is $\overline{E}^{\,A}=A\cap\overline{E}^{\,X}$, where the superscript specifies the space in which the closure is taken.

Proof. The set $A\cap\overline{E}^{\,X}$ is closed in $A$, by the closed-set characterization for subspaces, and it contains $E$ because $E\subseteq A$ and $E\subseteq\overline{E}^{\,X}$. Hence the closure of $E$ in $A$, which is the smallest closed subset of $A$ containing $E$, is contained in $A\cap\overline{E}^{\,X}$.

For the reverse inclusion, let $F$ be any closed subset of $A$ containing $E$. By the subspace closed-set characterization, $F=A\cap C$ for some closed $C\subseteq X$. Since $E\subseteq F\subseteq C$, the ambient closure $\overline{E}^{\,X}$ is contained in $C$. Consequently, $$ A\cap\overline{E}^{\,X}\subseteq A\cap C=F. $$ This holds for every closed $F\subseteq A$ containing $E$, so $A\cap\overline{E}^{\,X}$ is contained in the closure of $E$ in $A$. The two inclusions prove the formula. $\square$

Worked Example: The Same Set Has Different Closures in Different Spaces

Let $E=(0,1)$, viewed first as a subset of $\mathbb{R}$ and then as a subset of $A=(0,1]$. Its ambient closure in $\mathbb{R}$ is $[0,1]$. The closure formula gives $$ \overline{E}^{\,A}=A\cap\overline{E}^{\,\mathbb{R}} =(0,1]\cap[0,1]=(0,1]. $$ Thus $E$ is dense in $A$, since its closure in $A$ is all of $A$. In $\mathbb{R}$, however, its closure also contains $0$, which is not a point of $A$. The difference comes from restricting the closure to the space under consideration.

A Common Pitfall: Relative Interior

It is tempting to assume that the interior of $E$ in a subspace $A$ is always $A\cap\operatorname{int}_X(E)$. This is false in general: a point can have an open neighborhood relative to $A$ even when it has no ambient open neighborhood contained in $E$.

For instance, let $A=[0,1]$ and $E=[0,1/2)$. The point $0$ is an interior point of $E$ in $A$, because $$ A\cap(-1,1/2)=[0,1/2)\subseteq E. $$ But $0\notin\operatorname{int}_{\mathbb{R}}(E)$: every open interval around $0$ contains negative numbers, which are not in $E$. In fact, $\operatorname{int}_{\mathbb{R}}(E)=(0,1/2)$, so $A\cap\operatorname{int}_{\mathbb{R}}(E)=(0,1/2)$, which omits the relative interior point $0$.

The reliable test is always the subspace definition: search for an ambient open set whose intersection with $A$ is the set in question. Likewise, for relative closedness, use an ambient closed set and take its intersection with $A$. Keeping the ambient space explicit prevents the most common errors in subspace arguments.

Check Your Understanding

Use the subspace definition and the results above to answer the following questions.

What is the defining form of an open set in a subspace $A\subseteq X$?
Why is $[0,1/2)$ open in $[0,1]$ with the usual subspace topology, even though it is not open in $\mathbb{R}$?
Give an ambient closed set $C\subseteq\mathbb{R}$ such that $A\cap C=(0,1]$, where $A=(0,2)$.
If $B\subseteq A\subseteq X$, what does the transitivity theorem say about the topology on $B$?
For $E\subseteq A\subseteq X$, how is the closure of $E$ in $A$ calculated from its closure in $X$?
Why can relative interior contain a point that is not in the ambient interior?

What You'll Learn