Construct a topological space \(X\) by starting with \(\mathbb R\) with the usual topology and defining \(x\) to be equivalent to \(y\) if \(x-y\) is rational. Show that the resulting quotient or identification space \(X\) has the indiscrete topology, that is, the only open sets are \(\varnothing\) and \(X\).
Solution
Let \[ q:\mathbb R\longrightarrow X=\mathbb R/\!\sim \] be the quotient map.
1. Every nonempty open interval in \(\mathbb R\) contains a rational number.
Let \(a<b\). Choose \(n\in\mathbb N\) such that \[ n(b-a)>1. \] Let \[ m=\lfloor na\rfloor+1. \] Then \[ na<m\le na+1<nb. \] Dividing by \(n>0\) gives \[ a<\frac mn<b. \] Since \(m/n\in\mathbb Q\), the interval \((a,b)\) contains a rational number.
2. If \(W\subseteq X\) is open, then\[ U=q^{-1}(W) \]is an open saturated subset of \(\mathbb R\).
By the definition of the quotient topology, \(q^{-1}(W)\) is open in \(\mathbb R\). It is saturated because if \(x\in U\) and \(y\sim x\), then \[ q(y)=q(x)\in W, \] so \(y\in q^{-1}(W)=U\).
3. Every nonempty open saturated subset \(U\subseteq\mathbb R\) is all of \(\mathbb R\).
Let \(U\) be nonempty, open, and saturated. Choose \(x\in U\). Since \(U\) is open, there are real numbers \(a<b\) such that \[ x\in(a,b)\subseteq U. \]
Take an arbitrary \(y\in\mathbb R\). The interval \[ (y-b,y-a) \] is nonempty because \(a<b\). By 1, choose \[ r\in\mathbb Q\cap(y-b,y-a). \] The inequalities \[ y-b<r<y-a \] are equivalent to \[ a<y-r<b. \] Hence \[ y-r\in(a,b)\subseteq U. \] Moreover, \[ y-(y-r)=r\in\mathbb Q, \] so \[ y\sim y-r. \] Since \(U\) is saturated and \(y-r\in U\), it follows that \(y\in U\). The point \(y\) was arbitrary, so \(U=\mathbb R\).
4. The only open subsets of \(X\) are \(\varnothing\) and \(X\).
Both \(\varnothing\) and \(X\) are open in every topology.
Conversely, let \(W\subseteq X\) be a nonempty open set. By surjectivity of \(q\), its inverse image \[ U=q^{-1}(W) \] is nonempty. By 2 it is open and saturated, so 3 gives \[ q^{-1}(W)=\mathbb R. \] Applying \(q\) and using surjectivity, \[ W=q(q^{-1}(W))=q(\mathbb R)=X. \] Thus every nonempty open subset is all of \(X\), so the quotient topology is indiscrete.