# Can $\left| \bar{X} \right|$ be greater than $\left| \mathscr{P}(X) \right|$?

Let $$X$$ be a a subset of a linear continuum (consider with the order topology). Can $$\left| \bar{X} \right|$$ be greater than $$\left| \mathscr{P}(X) \right|$$? I know it can be equal, (e.g. for $$\mathbb{Q}$$ in $$\mathbb{R}$$). But can it be greater than that? I couldn't find any example neither any proof discarding it. Even such sets exist, is there any upper bound to the cardinality? It'll be great if anyone can shed some light.

EDIT : A linear continuum $$X,<$$ is a totally ordered set with l.u.b. property and $$\forall x, y \in X$$ with $$x, $$\exists z \in X$$ such that $$x < z .

• @CameronWilliams Sorry, there was a typo. Edited. – SinTan1729 Nov 12 '19 at 17:38
• It also cannot be equal -- $\mathbb Q$ is countable but $\mathbb R$ is not. – Matthew Daly Nov 12 '19 at 17:38
• To clarify then, what is $\overline{X}$ specifically in this context? A closure of some sort probably, but which closure? Algebraic closure? Topological closure? Of which topology? – JMoravitz Nov 12 '19 at 17:39
• It is topological closure. With respect to the order topology. – SinTan1729 Nov 12 '19 at 17:41
• What is the difference between $X$ and $\bar X$, then? – Claudius Nov 12 '19 at 17:50

Let $$X$$ be a subset of a linear continuum $$Y$$.
You would always have $$|\overline{X}|\le|\mathscr{P}(X)|$$.

Indeed, for each $$y\in\overline{X}$$ either $$y\in X$$, or $$y\in\overline{((-\infty,y)\cap X)}$$ or $$y\in\overline{((y,\infty)\cap X)}$$. Let $$A=\{y\in\overline{X}:y\in\overline{((-\infty,y)\cap X)}\}$$. Clearly if $$y,z$$ are different elements of $$A$$ then $$((-\infty,y)\cap X)\neq((-\infty,z)\cap X)$$. (Indeed, if $$y then $$\overline{(-\infty,y)}\subseteq(-\infty,y]$$ and $$z\notin(-\infty,y]$$.) Thus, the map from $$A$$ into $$\mathscr{P}(X)$$ defined by $$y\mapsto(-\infty,y)\cap X$$ is injective, hence $$|A|\le|\mathscr{P}(X)|$$.

Similarly, if $$B=\{y\in\overline{X}:y\in\overline{((y,\infty)\cap X)}\}$$ then $$|B|\le|\mathscr{P}(X)|$$. Hence $$|\overline{X}|\le|X\cup A\cup B|\le|\mathscr{P}(X)|$$. (The above assumes that $$\overline{X}$$ is infinite, but of course the case when it is finite is trivial, since then $$\overline{X}=X$$.)

In the statement of the problem, it is assumed that $$X$$ is a subset of a linear continuum. Say the linear continuum is $$Y$$ (so $$X\subseteq Y$$). Completeness properties of $$Y$$ do no matter. What is only used is that $$Y$$ is a linearly ordered set (same as totally ordered set, every two elements are comparable) with the order topology (same as open interval topology). (There seems to have been some confusion in the comments, but $$X$$ does not denote the linear continuum itself, in the OP, $$X$$ is only a subset of the linear continuum.)