# Exercise 7.77#

Show that a subset $$π β β$$ is open in the subspace topology of $$β β πβ$$ iff $$π β© β$$ is open in the usual topology on $$β$$ defined in Example 7.26.

Every point in $$β_s$$ (for β as a subspace of πβ) is effectively an open ball in the sense of Ball (mathematics) (an open line, in this case). That is, we define it as $$[d,u] \in πβ$$ where $$d = u$$. That means the basic open sets are:

\begin{split} \begin{align} π_{[π,π]} &β \{[π,π’] β πβ \enspace | \enspace a < π β€ π’ < b\} \\ π_{[π,π]} &β \{[d,d] β πβ \enspace | \enspace a < π < b\} \end{align} \end{split}

Then $$π_{[π,π]}$$ 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.

## Authorβs solution#

A subset $$π β β$$ is open in the subspace topology of $$β β πβ$$ iff there is an open set $$π' β πβ$$ with $$π = π' β© β$$. We want to show that this is the case iff $$π$$ is open in the usual topology.

Suppose that $$π$$ is open in the subspace topology. Then $$π = π' β© β$$, where $$π' β πβ$$ is the union of some basic opens, $$π' = \bigcup_{πβπΌ} π_{[π_π,π_π]}$$, where $$π_{[π_π,π_π]} = \{[π,π’] β πβ \enspace | \enspace π_π < π < π’ < π_π\}$$. Since $$β = \{[π₯, π₯] β πβ\}$$, the intersection $$π' β© β$$ will then be:

$π = \bigcup_{i \in I} \{π₯ β β \enspace | \enspace π_π < π₯ < π_π\}$

and this is just the union of open balls $$π΅(π_π, π_π)$$ where $$π_π := \frac{π_π+π_1}{2}$$ is the midpoint and $$r_i = \frac{b_π-a_i}{2}$$ is the radius of the interval $$(π_π, π_π)$$. The open balls $$π΅(π_π, π_π)$$ are open in the usual topology on β and the union of opens is open, so π is open in the usual topology.

Suppose that $$π$$ is open in the usual topology. Then $$π = \bigcup_{j \in J} π΅(π_π, π_π)$$ for some set $$π½$$. Let $$π_π := π_π β π_π$$ and $$π_π := π_π + π_π$$. Then:

$π = \bigcup_{πβπ½} \{π₯ β β \enspace | \enspace π_π < π₯ < π_π \} = \bigcup_{πβπ½} (π_{[π_π, π_π]} β© β) = \left(Β­\bigcup_{πβπ½} π_{[π_π, π_π]}\right) β© β$

which is open in the subspace topology.