However we clearly have a x for every a x x which

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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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