# Is a subset of a topological space able to induce a quotient space

As far as I know, a quotient space of a topological space must be defined wrt an equivalence relation on the topological space.

1. But then I wonder what the equivalence relation is in the following example from Wikipedia:

In topology, especially algebraic topology, the cone $CX$ of a topological space $X$ is the quotient space: $$CX = (X \times I)/(X \times \{0\})\,$$ of the product of $X$ with the unit interval $I = [0, 1]$. Intuitively we make $X$ into a cylinder and collapse one end of the cylinder to a point.

2. Is a subset of a topological space able to induce an equivalence relation on the topological space, and to induce a quotient space? I know it is true for a subspace of a vector space.

Thanks and regards!

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Let $X$ be a topological space, and let $A\subseteq X$. $A$ defines the following equivalence relation $\overset{A}\sim$ on $X$: for $x,y\in X$, $x\overset{A}\sim y$ iff either $x=y$, or $\{x,y\}\subseteq A$. The quotient space $X/A$ is defined to be the same as the quotient space $X/\overset{A}\sim$.
Added: To put it a little differently, the partition of $X$ induced by $\overset{A}\sim$ is $$\{A\}\cup\big\{\{x\}:x\in X\setminus A\big\}\;.$$ The topology of of the quotient space is defined as follows: $V\in X/A$ is open iff $\pi^{-1}[V]$ is open in $X$, where $\pi:X\to X/A$ is the obvious quotient map.
THanks! (1) It seems that the equivalence classes are $A$, and the singleton subsets of $A^c$? (2) This definition doesn't depend on topology at all, actually not assuming any structure on a set $X$? (3) When $X$ is a vector space, and $A$ is a subspace of $X$, the quotient space in the vector space sense doesn't coincide with the definition in your reply. Does it? – Tim Apr 3 '12 at 22:58
@BrianM.Scott Q.3. in this note says that the quotient space $X/A =(X\setminus A)\coprod \{*\}$. I cant find the basic lecture and i guess $\coprod$ is disjoint union but its so complicated for me to show that $X/A$ and $(X\setminus A)\coprod \{*\}$ are equal as topological spaces and then how we can clarify the open sets. could you help me, please? – M.A. Apr 29 at 15:44
@M.A.: Yes, that’s the disjoint union. That description corresponds to the partition in the Added part of my answer: $\{*\}$ represents the piece $A$ of the partition, and each point $x\in X\setminus A$ corresponds to the piece $\{x\}$ of the partition. For the topology, let $U$ be an open set in $X$. If $U\cap A=\varnothing$, then $U$ is open in $X/A$. If $U\supseteq A$, then $U/A$ is open in $X/A$, where $U/A=(U\setminus A)\cup\{*\}$, the set obtained by collapsing $A$ to a single point. Every open set in $X/A$ is of one of these forms. – Brian M. Scott Apr 29 at 22:13