Collapsing a Subspace to One Point

What Does It Mean to Collapse a Subspace?

Collapsing a subset $A$ of a space $X$ means treating all points of $A$ as a single point while keeping each point outside $A$ distinct. The quotient set is straightforward to describe; its topology takes a little more care. The key test is to pull a proposed set back to $X$ and check whether that preimage is open.

Throughout, $X$ has topology $\mathcal{T}$, and $A\subseteq X$ is nonempty. Define a relation on $X$ by declaring $x\sim y$ if either $x=y$, or both $x\in A$ and $y\in A$. The quotient set $X/{\sim}$ consists of the equivalence classes, and $q:X\to X/{\sim}$ sends each point to its class. Write $*=A$ for the class containing all of $A$, and write $[x]$ for the class of a point $x\notin A$.

Definition: The quotient topology on $X/{\sim}$ is the topology in which $W\subseteq X/{\sim}$ is open if and only if $q^{-1}(W)$ is open in $X$. For a subset $U\subseteq X$, its saturation is $q^{-1}(q(U))$, the union of all equivalence classes that meet $U$.

The equivalence classes have a simple description: there is one class $A$, and every point of $X\setminus A$ forms a singleton class. Indeed, every point of $A$ is related to every other point of $A$; no point outside $A$ is related to a different point; and a point outside $A$ is not related to any point of $A$. The relation is reflexive, symmetric and transitive: equality handles any comparison involving a point outside $A$, while points of $A$ all belong to the same class. Thus it is an equivalence relation.

The Saturation Test

Theorem: Let $A\ne\varnothing$, let $q:X\to X/{\sim}$ collapse $A$, and let $U\subseteq X$. Then $$ q^{-1}(q(U))= \begin{cases} U, & U\cap A=\varnothing,\\ U\cup A, & U\cap A\ne\varnothing. \end{cases} $$ Consequently, $q(U)$ is open in $X/{\sim}$ if and only if the set on the right is open in $X$.

Proof. A point $x$ lies in $q^{-1}(q(U))$ exactly when its equivalence class meets $U$. If $U\cap A=\varnothing$, no point of $U$ belongs to the class $A$, and every other class is a singleton. Therefore precisely the points of $U$ have classes meeting $U$, giving $q^{-1}(q(U))=U$. If $U\cap A\ne\varnothing$, the class $A$ meets $U$, so every point of $A$ belongs to the preimage; every point outside $A$ belongs to it exactly when it is in $U$. Hence the preimage is $U\cup A$. By the definition of the quotient topology, $q(U)$ is open exactly when its preimage is open. $\square$

This test includes an important distinction. If $U$ misses $A$, its image has preimage $U$, not $U\cup A$. If $U$ meets even one point of $A$, its image includes the collapsed point $*$, and its preimage must include all of $A$. Thus openness of $U$ alone does not settle whether $q(U)$ is open.

Worked Example: Listing the Classes in an Interval

Give $I=[0,1]$ the subspace topology inherited from the standard topology on $\mathbb{R}$. Let $A=[1/3,2/3]$, and collapse $A$ to one point.

The quotient set has the class $*=[1/3,2/3]$, and singleton classes $[x]=\{x\}$ for every $x\in[0,1/3)\cup(2/3,1]$. In particular, $1/3$ and $2/3$ belong to $*$, while $0$ and $1$ remain separate singleton classes. The quotient map satisfies $q(x)=*$ for $1/3\le x\le2/3$, and $q(x)=[x]$ otherwise.

Check: the endpoints are included in $A$, so $q(1/3)=q(2/3)=*$. Since neither $0$ nor $1$ lies in $A$, $q(0)=[0]$ and $q(1)=[1]$. No two distinct points outside $A$ are identified.

Worked Example: An Open Image That Contains the Collapsed Point

Continue with $I=[0,1]$ and $A=[1/3,2/3]$. Take $U=[0,3/4)$, which is open in $I$, since $U=I\cap(-1,3/4)$. It contains all of $A$, so $q^{-1}(q(U))=U\cup A=U$, an open subset of $I$. Therefore $q(U)$ is open in the quotient.

More explicitly, $q(U)$ consists of the point $*$, the classes $[x]$ for $0\le x<1/3$, and the classes $[x]$ for $2/3<x<3/4$. The endpoints $1/3$ and $2/3$ are not listed separately because both are already in $*$. Since the preimage of this set is exactly $[0,3/4)$, the quotient-open test is satisfied.

Check: $3/4\notin U$, and $3/4\notin A$, so its class $[3/4]$ is not in $q(U)$. Every point of $A$ is in $U$, so the only points of $U$ that are not represented by $*$ are exactly the singleton classes listed above.

Worked Example: An Open Set Whose Image Is Not Open

Again let $I=[0,1]$ and $A=[1/3,2/3]$, but now set $B=(0,1/2)$. This is open in $I$, and it meets $A$ because, for example, $2/5\in B\cap A$: $0<2/5<1/2$ and $1/3\le2/5\le2/3$. Since $B$ meets $A$, its saturation is $$ q^{-1}(q(B))=B\cup A=(0,2/3]. $$

This saturation is not open in $I$. It contains $2/3$, but every open neighborhood of $2/3$ in $I$ contains points greater than $2/3$. Those points are not in $(0,2/3]$, so no such neighborhood is contained in the saturation. The quotient-open test now gives that $q(B)$ is not open.

Check: the quotient image is $q(B)=\{*\}\cup\{[x]:0<x<1/3\}$. Its preimage includes all of $A$, including $2/3$, because $B$ meets $A$. It also includes the points of $B$ below $1/3$, and it contains no point greater than $2/3$. These are exactly the points of $(0,2/3]$.

The Same Test in a Disk

Let $D^2=\{x\in\mathbb{R}^2:\|x\|\le1\}$ have the subspace topology inherited from the Euclidean topology on $\mathbb{R}^2$. Choose the closed subdisk $A=\{x\in D^2:\|x\|\le1/2\}$, and collapse it to $*$. Each point of $D^2\setminus A$, equivalently each point with norm greater than $1/2$, remains a singleton class. The saturation test applies exactly as it did for the interval.

Worked Example: A Quotient-Open Set in the Disk

Take $V=\{x\in D^2:\|x\|<3/4\}$. This set is open in $D^2$, since it is the intersection of $D^2$ with the open Euclidean ball of radius $3/4$ centered at the origin. It contains the whole collapsed subdisk: if $x\in A$, then $\|x\|\le1/2<3/4$. Thus $q^{-1}(q(V))=V\cup A=V$, so $q(V)$ is open in the quotient.

Its quotient image consists of $*$ and the singleton classes of points $x$ with $1/2<\|x\|<3/4$. The strict inequalities matter: norm $1/2$ is in the collapsed subdisk, and norm $3/4$ is not in $V$. The preimage of this described image is precisely the disk of radius less than $3/4$, which verifies the claim.

Worked Example: A Nonopen Image in the Disk

Let $U=\{x\in D^2:x_1>0\}$, where $x_1$ is the first coordinate. This is open in $D^2$, being the intersection with the open half-plane $x_1>0$. It meets $A$, for instance at $(1/4,0)$, whose norm is $1/4\le1/2$ and whose first coordinate is positive. Its saturation is $U\cup A$.

To show that $U\cup A$ is not open, consider $p=(-1/2,0)\in A$, and for $n\ge3$ let $p_n=(-1/2-1/n,0)$. Then $$ \|p_n\|=1/2+1/n\le5/6<1, $$ so $p_n\in D^2$, and $\|p_n\|>1/2$, so $p_n\notin A$. Its first coordinate is negative, so $p_n\notin U$. Therefore $p_n\notin U\cup A$, even though $p\in U\cup A$ and $p_n\to p$.

Every open neighborhood of $p$ in $D^2$ contains all sufficiently late terms of this convergent sequence, so no such neighborhood is contained in $U\cup A$. Hence the saturation is not open, and $q(U)$ is not open. Check: the inequality above uses $n\ge3$, so $1/2+1/n\le1/2+1/3=5/6<1$; the points used really do lie in the disk.

A Common Mistake and the General Open-Set Criterion

Common pitfall. When testing whether $q(B)$ contains the collapsed point, do not replace an open interval by its closure. For $B=(0,1/3)$ and $A=[1/3,2/3]$, the endpoint $1/3$ is not in $B$. Thus $B\cap A=\varnothing$, and $$ q^{-1}(q(B))=B=(0,1/3). $$ This is open in $I$, so $q(B)$ is open. The incorrect calculation $q^{-1}(q(B))=(0,2/3]$ would wrongly include $A$ even though $B$ does not meet it.

To make the contrasting calculation valid, use $B=(0,1/2)$. Now $1/3\in B\cap A$, so the saturation really is $B\cup A=(0,2/3]$, which is not open in $I$. The criterion depends on actual intersection, including whether an endpoint is included.

Takeaway: A subset $W$ of the quotient is open exactly when its full preimage $q^{-1}(W)$ is open in $X$. For a set presented as $q(U)$, first compute its saturation: it is $U$ if $U$ misses $A$, and $U\cup A$ if $U$ meets $A$.

This is why the quotient topology is useful: it specifies the topology after identification without relying on a picture of what the result should look like. A quotient-open set is tested in the original space, where openness is already understood. The collapsed class also explains why a small piece of $A$ cannot be treated as though only that piece had been identified: once an image includes $*$, its preimage must contain every point of $A$.

Check Your Understanding

For each question, write down the relevant equivalence classes or preimage before deciding.

In $I=[0,1]$, collapse $A=[1/4,3/4]$. Describe the class of $0.6$, the class of $1$, and the class $*$.
With the same $I$ and $A$, let $U=(0,1/5)$. Find $q^{-1}(q(U))$ and decide whether $q(U)$ is open.
With the same $I$ and $A$, let $U=(1/5,1/2)$. Find its saturation and decide whether its quotient image is open.
For the disk example, explain why the set $\{x\in D^2:\|x\|<3/4\}$ contains all of the collapsed subdisk.
State the quotient-open test for a subset $W\subseteq X/{\sim}$, and distinguish it from the saturation test for a set written as $q(U)$.