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ID: past-exam-of-the-graph-theory-and-probabilistic-combinatorics-course-of-the-university-of-oxford/2010/a/2/c/solution

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Past exam of the graph Theory and Probabilistic Combinatorics course of the University of Oxford / 2010 / A / 2 / c / Solution by

Codex  0 2026-10-31

No. Colour the positive integers by alternating dyadic blocks:

C_{0} = ⋃_{j \geq 0} ([2^{2 j}, 2^{2 j + 1}) \cap N), C_{1} = ⋃_{j \geq 0} ([2^{2 j + 1}, 2^{2 j + 2}) \cap N) .

Every positive integer

x

and its double

2 x

have opposite colours. A nonempty set closed under addition contains both

x

and

x + x = 2 x

, so no such set can lie in one class. In particular there is no infinite monochromatic additive subsemigroup.

The suggested arithmetic-progression argument gives another way to see the obstruction. If

a \in S

and

S

is closed under addition, then

a, 2 a, 3 a, \dots

all lie in

S

. But every infinite arithmetic progression of positive step

d

meets every sufficiently long dyadic block, because a block of at least

d

consecutive integers contains a representative of the progression's residue class. Both colours contain arbitrarily long blocks, so neither colour contains an infinite arithmetic progression.

Figure 1.
Alternating dyadic colouring and the doubling obstruction
. Every positive integer and its double lie in opposite coloured blocks. Details.

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