Is it true that no linear order can have CCC?

Question

I am reading about countable chain condition on Wikipedia https://en.wikipedia.org/wiki/Countable_chain_condition

I want to quickly verify that if $(X, \leq)$ is a linear order, then it cannot possibly have the countable chain condition. Is this correct?

Because in a linear order, every elements are comparable. So there does not exist any anti-chain, let alone a countable anti-chain.

Thanks!

Answer 1

The countable chain condition means that there does not exist any uncountable antichain. It doesn't mean that there must exist a countably infinite antichain. So in fact, every linear order does have the countable chain condition, by your argument.

(Beware, though, that when talking about linear orders, "countable chain condition" usually refers to a different condition. Namely, it refers to the countable chain condition on the poset of non-empty open intervals in the linear order, ordered by inclusion. So a linear order is said to be ccc if it has no uncountable collection of disjoint open intervals.)