10.3 Quotients[2C3]

E288

[0R2] Suppose that \(d\) satisfies all distance requirements except ”separation property”. Consider the relation \(∼\) on \(X\) defined as \(x∼ y\iff d(x,y)=0\); show that is an equivalence relation. Let’s define \(Y=X/∼\); show that the function \(d\) “passes to the quotient”, that is, there exists \(\tilde d:Y× Y→ [0,∞)\) such that, for every choice of classes \(s,t∈ Y\) and every choice of \(x∈ s,y∈ t\) you have \(\tilde d(s,t)=d(x,y)\). Finally, show that \(\tilde d\) is a distance on \(Y\).

This procedure is the metric space equivalent of Kolmogoroff quotient.

E288

[0R3] Let \((X,d)\) be a metric space and \(∼\) an equivalence relation on \(X\). Let \(Y=X/∼\) be the quotient space. We define the function \(𝛿:Y^ 2→ℝ\) as

\begin{equation} 𝛿(x,y) = \inf \{ d(s,t) : s∈ x, t∈ y \} ~ ~ .\label{eq:dist_ quoziente} \end{equation}

289

Is it a distance on \(Y\)? Which properties does it enjoy among those indicated in 275? Hidden solution: [UNACCESSIBLE UUID ’0R4’]

E288

[0R5] Let \((X,d)\) be a metric space where \(X\) is also a group. Let \(Θ\) be a subgroup.

We define that \(x∼ y\iff x y^{-1}∈Θ\). It is easy to verify that this is an equivalence relation. Let \(Y=X/∼\) be the quotient space. ¹

Suppose \(d\) is invariant with respect to left multiplication by elements of \(Θ\):

\begin{equation} d(x,y)=d(𝜃 x,𝜃 y)~ ~ ∀ x,y∈ X,∀ 𝜃∈Θ~ ~ . \label{eq:d_ invarian_ grupp} \end{equation}

290

(This is equivalent to saying that, for every fixed \(𝜃∈Θ\) the map \(x↦ 𝜃 x\) is an isometry). We define the function \(𝛿:Y^ 2→ℝ\) as in ??.

• Show that, taken \(s,t∈ X\),

  \begin{equation} 𝛿([s],[t]) = \inf \{ d(s,𝜃 t) : 𝜃 ∈ Θ \} \label{eq:dist_ quoz_ 1gruppo} \end{equation}

  291

  where \([s]\) is the class of elements equivalent to \(s\).

• Show that \(𝛿≥ 0\), that \(𝛿\) is symmetric and that \(𝛿\) satisfies the triangle inequality.

• Suppose that, for every fixed \(t∈ X\), the map \(𝜃↦ 𝜃 t\) is continuous from \(Θ\) to \(X\); suppose also that \(Θ\) is closed: then \(𝛿\) is a distance. ²

Hidden solution: [UNACCESSIBLE UUID ’0R6’]