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I try to solve this problem in General Topology Stephen Willard :-

Show that

  1. $|βN|\ge|βQ|$. [Consider any one-one map of $N$ onto $Q$ and use Theorem]

  2. $|βQ| \ge|βR|$. [Consider the inclusion map of $Q$ onto $R$ and use Theorem]

  3. $|βN| =|βQ| = |βR| = 2^\mathfrak{c}$. [N is C*-embedded in R ]

Definition 1: $βΧ$ is the Stone-Cech compactification of X.

Definition 2: A subset $A$ of a space $Τ$ is $C*$-embedded in $Τ$ iff every bounded continuous real-valued function on $A$ can be extended to $T$.

Theorem. If К is a compact Hausdorff space and $f: X \to K$ is
continuous, there is a continuous $F: βΧ \to K$ such that $F \circ e = f $

Is there exist a one-one map of $N$ onto $Q$ ? Which one ?

If there is one, what should i do then ??

Hamada Al
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  • https://math.stackexchange.com/questions/2492057/stone-cech-compactification-cardinality/2492092#2492092 try to take a look also to this question – Soma Oct 27 '17 at 19:58
  • Could you give me any one-to-one map of N onto Q ?? – Hamada Al Oct 27 '17 at 20:21
  • Here many examples: https://math.stackexchange.com/questions/659302/how-to-prove-that-mathbbq-the-rationals-is-a-countable-set – Soma Oct 27 '17 at 20:29
  • If $X$ is a normal space and $Y$ is closed in $X$ then the closure of $Y$ in $\beta X$ is $\beta Y.$....and hence $|\beta Y|\leq |\beta X|$..... Let $X=R$ and $Y=N.$ – DanielWainfleet Oct 29 '17 at 09:04

1 Answers1

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I answered the first two in this answer For the last you only need that $|\beta \mathbb{R}| \ge |\beta \mathbb{N}|$ which is easy to see:

$X= \overline{\mathbb{N}}$( as a subset of $\beta \mathbb{R}$) is compact and has $\mathbb{N}$ as a dense subset and if $f: \mathbb{N} \to [0,1]$ is continuous, then we can extend $f$ to a map $f'$ from $\mathbb{R}$ to $[0,1]$ as $\mathbb{N}$ is $C^\ast$-embedded in the reals. Then we extend $f'$ to $\beta f' : \beta \mathbb{R} \to [0,1]$ and note that of course $\beta f' |_X$ is an extension of $f$ to $X$. So $X$ obeys the extension property that uniquely characterises $\beta \mathbb{N}$ so that $X =\overline{\mathbb{N}}^{(\beta \mathbb{R})} \simeq \beta \mathbb{N}$ and as $X \subseteq \beta \mathbb{R}$ we clearly have $|\beta \mathbb{N}| \le |\beta \mathbb{R}|$ as required.

In fact the above shows that if $A$ is $C^\ast$-embedded in $X$ then $\beta A \hookrightarrow \beta X$ (as $\overline{A}^{(\beta X)}$ in fact).

I believe that $|\beta \mathbb{N}|=2^{\mathfrak{c}}$ is shown in the text itself.

Henno Brandsma
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