If \(\mathcal{T}\) and \(\mathcal{T}'\) are topologies on \(X\) and \(\mathcal{T}'\) is strictly finer than \(\mathcal{T}\), what can you say about the corresponding subspace topologies on the subset \(Y\) of \(X\)?
Solution
Let \(\mathcal T_Y\) and \(\mathcal T'_Y\) be the subspace topologies induced by \(\mathcal T\) and \(\mathcal T'\). Since \(\mathcal T\subseteq\mathcal T'\), every set \(Y\cap U\) with \(U\in\mathcal T\) is also of the form \(Y\cap U\) with \(U\in\mathcal T'\). Hence \[ \mathcal T_Y\subseteq\mathcal T'_Y. \] The inclusion need not be strict even when \(\mathcal T\subsetneq\mathcal T'\). For example, let \[ X=\{a,b\},\quad \mathcal T=\{\varnothing,X\},\quad \mathcal T'=\{\varnothing,\{a\},X\},\quad Y=\{b\}. \] Both induced topologies on the singleton \(Y\) are \(\{\varnothing,Y\}\). Thus refinement passes to every subspace, but strict refinement need not.