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Exercise 5.2.3
If a linearly ordered set has a countable dense subset, then .
Answers
Note that the countable dense subset is dense in and not just in itself as explained in the errata list.
Proof. Suppose that is our linearly ordered set and is the countable dense subset of . We construct an by defining
for any . Clearly for such we have that so that .
Now we claim that is injective. So consider any such that . Without loss of generality we can assume that . Since is dense in there is a where . From this it follows that but that since it is not true that . Hence clearly so that we have shown that is injective.
Thus we have
where we have used Theorem 5.1.9. □