Exercise 5.2.3

If a linearly ordered set P has a countable dense subset, then | P | 2 0 .

Answers

Note that the countable dense subset is dense in P and not just in itself as explained in the errata list.

Proof. Suppose that ( P , < ) is our linearly ordered set and R is the countable dense subset of P . We construct an f : P 𝒫 ( R ) by defining

f ( x ) = { y R y < x }

for any x P . Clearly for such x we have that f ( x ) R so that f ( x ) 𝒫 ( R ) .

Now we claim that f is injective. So consider any x , y P such that x y . Without loss of generality we can assume that x < y . Since R is dense in P there is a z R where x < z < y . From this it follows that z f ( y ) but that z f ( x ) since it is not true that z < x . Hence clearly f ( x ) f ( y ) so that we have shown that f is injective.

Thus we have

| P | | 𝒫 ( R ) | = 2 | R | = 2 0 ,

where we have used Theorem 5.1.9. □

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2024-07-15 11:42
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