# Ordered set of cardinality continuum that has more than continuum many initial segments

The book I am reading has the following has the following exericse: "Show that there is an ordered set of cardinality continuum that has more than continuum many initial segments," but I don't see how this plays out exactly...thought maybe someone could shed some light.

It is enough to show an ordered set of cardinality bigger than an infinite cardinal κ, and a dense subset A ⊂ B in it of cardinality κ. In fact, then every element b ∈ B determines the initial segment $S_{b}$ = {a ∈ A : a ≺ b} of , hence has more than κ initial segments (note that for $b_1$ is $\neq$ $b_2$ the initial segments $S_{b_1}$ and $S_{b_2}$ are different by the density of A in B).

• What's the definition of an initial segment? Is it a set of the form $S(x) = \{ y : y < x\}$, or is it a set $S$ with the property $\bigl(\forall x,y\bigr)\bigl( x\in S \land y < x \implies y\in S\bigr)$? – Daniel Fischer Feb 14 '16 at 16:24
• The second definition. – Learner Feb 14 '16 at 16:29
• Okay. Then you know an example with smaller cardinalities, $\mathbb{Q}$, which is countable, has continuum many initial segments. Can you use that to guide you? – Daniel Fischer Feb 14 '16 at 16:32
• Is this under something like the Continuum Hypothesis? – Asaf Karagila Feb 14 '16 at 16:42
• It might be helpful to name the book you are reading (author, title, etc.), as well as say where inside the book this exercise can be found. – i'm nobody who are you Feb 14 '16 at 18:24

Let $\kappa$ be the least cardinal such that $2 ^{\kappa}$ is strictly larger than continuum. Then the lexicographic ordering of $2^{< \kappa}$ is such an order. I will leave the details to you.