Exercise 7.77

Show that a subset \(𝑈 ⊆ ℝ\) is open in the subspace topology of \(ℝ ⊆ 𝕀ℝ\) iff \(𝑈 ∩ ℝ\) is open in the usual topology on \(ℝ\) defined in Example 7.26.

Alternative answer

Every point in \(ℝ_s\) (for ℝ as a subspace of 𝕀ℝ) is effectively an open ball in the sense of Ball (mathematics) (an open line, in this case). That is, we define it as \([d,u] \in 𝕀ℝ\) where \(d = u\). That means the basic open sets are:

\[\begin{split} \begin{align} 𝑜_{[𝑎,𝑏]} &≔ \{[𝑑,𝑢] ⊆ 𝕀ℝ \enspace | \enspace a < 𝑑 ≤ 𝑢 < b\} \\ 𝑜_{[𝑎,𝑏]} &≔ \{[d,d] ⊆ 𝕀ℝ \enspace | \enspace a < 𝑑 < b\} \end{align} \end{split}\]

Then \(𝑜_{[𝑎,𝑏]}\) is equivalent to the open line \((a,b)\); we can take \(d\) to be any point between \(a\) and \(b\). Since these two topological spaces share an essentially equivalent basis, they should be otherwise equivalent.

Author’s solution

A subset \(𝑈 ⊆ ℝ\) is open in the subspace topology of \(ℝ ⊆ 𝕀ℝ\) iff there is an open set \(𝑈' ⊆ 𝕀ℝ\) with \(𝑈 = 𝑈' ∩ ℝ\). We want to show that this is the case iff \(𝑈\) is open in the usual topology.

Suppose that \(𝑈\) is open in the subspace topology. Then \(𝑈 = 𝑈' ∩ ℝ\), where \(𝑈' ⊆ 𝕀ℝ\) is the union of some basic opens, \(𝑈' = \bigcup_{𝑖∈𝐼} 𝑜_{[𝑎_𝑖,𝑏_𝑖]}\), where \(𝑜_{[𝑎_𝑖,𝑏_𝑖]} = \{[𝑑,𝑢] ∈ 𝕀ℝ \enspace | \enspace 𝑎_𝑖 < 𝑑 < 𝑢 < 𝑏_𝑖\}\). Since \(ℝ = \{[𝑥, 𝑥] ∈ 𝕀ℝ\}\), the intersection \(𝑈' ∩ ℝ\) will then be:

\[ 𝑈 = \bigcup_{i \in I} \{𝑥 ∈ ℝ \enspace | \enspace 𝑎_𝑖 < 𝑥 < 𝑏_𝑖\} \]

and this is just the union of open balls \(𝐵(𝑚_𝑖, 𝑟_𝑖)\) where \(𝑚_𝑖 := \frac{𝑎_𝑖+𝑏_1}{2}\) is the midpoint and \(r_i = \frac{b_𝑖-a_i}{2}\) is the radius of the interval \((𝑎_𝑖, 𝑏_𝑖)\). The open balls \(𝐵(𝑚_𝑖, 𝑟_𝑖)\) are open in the usual topology on ℝ and the union of opens is open, so 𝑈 is open in the usual topology.

Suppose that \(𝑈\) is open in the usual topology. Then \(𝑈 = \bigcup_{j \in J} 𝐵(𝑚_𝑗, 𝜖_𝑗)\) for some set \(𝐽\). Let \(𝑎_𝑗 := 𝑚_𝑗 − 𝜖_𝑗\) and \(𝑏_𝑗 := 𝑚_𝑗 + 𝜖_𝑗\). Then:

\[ 𝑈 = \bigcup_{𝑗∈𝐽} \{𝑥 ∈ ℝ \enspace | \enspace 𝑎_𝑗 < 𝑥 < 𝑏_𝑗 \} = \bigcup_{𝑗∈𝐽} (𝑜_{[𝑎_𝑗, 𝑏_𝑗]} ∩ ℝ) = \left(­\bigcup_{𝑗∈𝐽} 𝑜_{[𝑎_𝑗, 𝑏_𝑗]}\right) ∩ ℝ \]

which is open in the subspace topology.