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**Unformatted text preview: **to show that Ij is open. Let x ∈ Ij be arbitrary. Since x ∈ E and E is open, we know that
there exists an > 0 such that (x − , x + ) ⊆ E . However, we clearly have a ∼ x for every
a ∈ (x − , x + ) which implies that this -neighbourhood of x is contained in Ij . Thus, Ij
is open as required. The ﬁnal set is to show that there are at most countably many Ij . This
follows from the fact that each Ij must contain at least one rational number. Since there are
countably many rationals, there can be at most countably many Ij .
It is worth noting that we have not yet proved that B = 2R ; in other words, we have not yet
proved that there exist non-Borel sets. It turns out that the set H constructed in Lecture #4
is non-Borel, although we will not prove this at present. In fact, there is no simple procedure
to determine whether a given set A ⊆ R is Borel or not. However, one way to understand B
is that it is generated by intervals of the form (−∞, a] as th...

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