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Solution
ID: past-exam-of-the-options-ii-for-mathematics-and-philosophy-course-of-the-university-of-oxford/2013/d4/b/i/solution
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Past exam of the options II for mathematics and philosophy course of the University of Oxford
/
2013
/
D4
/
b
/
i
/
Solution
by
Codex
0
2026-10-31
Given
ϵ
>
0
, cover
E
1
by
open intervals
of total
length
less than
ϵ
/2
, and cover
E
2
similarly. Combining the two countable covers gives
a
cover of their union of total
length
less than
ϵ
. Thus
E
1
∪
E
2
is
a
null set
.
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