Question on a proof from Jech's Set Theory of: If $X$ is a set of cardinals, then $\sup X$ is a cardinal.

Acid2 10/14/2018 at 07:27. 3 answers, 0 views
set-theory cardinals ordinals

The proof: Let $\alpha= \sup X$. If $f$ is a bijective mapping of $\alpha$ onto some $\beta < \alpha$, let $\kappa \in X$ be such that $\beta < \kappa \le \alpha$. Then $|\kappa|=|\{f(\xi): \xi < \kappa\}| \le \beta$, a contradiction. Thus, $\alpha$ is a cardinal.

*An ordinal $\alpha$ is a cardinal if $|\alpha|\neq |\beta|$ for all $\beta < \alpha$.


Here's how far I think I understand the logic of the proof; first we assume $\alpha$ isn't a cardinal. This means there would be a $\beta$ less than $\alpha$ which is equinumerous to $\alpha$. We then suppose another ordinal $\kappa$ greater than $\beta$ but that injects into $\alpha$. What I don't get is how do we know that $|\kappa|=|\{f(\xi): \xi < \kappa\}|$, not to mention how $|\{f(\xi): \xi < \kappa\}| \le \beta$. If all of that stands true, then I get that we arrive at the contradiction; $\beta < \kappa$ means $\beta \in \kappa$ and that implies there can't be an injection from $\kappa$ into $\beta$ ($\kappa \le \beta$).

3 Answers


user642796 09/07/2012.

(I will write $X \approx Y$ to denote that $X$ and $Y$ are equinumerous.)

Note the following simple fact: If $f : X \to Y$ is a bijection, then for every $A \subseteq X$ the restriction $f \restriction A$ is a bijection between $A$ and $f''A$, which gives us that $A \approx f''A$ for all $A \subseteq X$.

In the context of the proof above, we have that $\kappa \approx f''\kappa = \{ f(\xi) : \xi \in \kappa \}$. As $f(\xi) \in \beta$ for all $\xi \in \alpha$, then surely $f''\kappa \subseteq \beta$. We thus have that $f''\kappa \subseteq \beta \subseteq \kappa$, and so by Cantor–Bernstein–Schroeder we may conclude that $f''\kappa \approx \beta \approx \kappa$. This contradicts the fact that $\kappa$ is a cardinal!


I would have proceeded slightly differently and avoided the troubling function $f$.

If $\alpha = \sup (X)$ is not a cardinal, then $|\alpha| < \alpha$ (where $|\alpha|$ denotes the unique cardinal equinumerous to $\alpha$). As $|\alpha|$ cannot be an upper bound for $X$, there is a cardinal $\kappa \in X$ such that $|\alpha| < \kappa \leq \alpha$. Note that by Cantor–Bernstein–Schroeder (as $|\alpha| \subseteq \kappa \subseteq \alpha$) it follows that $|\alpha| \approx \kappa$, contradicting that $\kappa$ is a cardinal!


William 10/31/2012.

Let $X$ be a set of cardinals. Let $\alpha = \sup X$. Suppose that $\sup X$ is not a cardinal. Then there exists a $\beta < \alpha$ such that there exists a bijection $f : \alpha \rightarrow \beta$. Since $\alpha$ is not a cardinal and $X$ is a set of cardinals, you have that $\alpha \notin X$. Since $\alpha = \sup X$, you must have that there exists a cardinal $\gamma \in X$ such that $\beta < \gamma < \alpha$. Thus $f|\gamma$ ($f$ restricted to $\gamma$) is a injection from $\gamma$ to $\beta$. By definition of $\beta < \gamma$ (as ordinals), you have that injection from $\beta$ into $\gamma$. By the Shoder-Bernstein Theorem, there is a bijection between $\gamma$ and $\beta$. But $\gamma \in X$ is a cardinal; hence, there can not be a bijection between $\gamma$ and the smaller ordinal $\beta$. Contradiction!


martini 09/07/2012.

As $\alpha$ isn't a cardinal and $\kappa \in X$ is one, we must have $\kappa < \alpha$. As $f$ is bijective, we have \[ \kappa = |\kappa| = |f''\kappa| = |\{f(\xi) \mid \xi < \kappa\}| \le |\{f(\xi) \mid \xi < \alpha \}| = |\beta| < \kappa. \]


HighResolutionMusic.com - Download Hi-Res Songs

1 Alan Walker

Diamond Heart flac

Alan Walker. 2018. Writer: Alan Walker;Sophia Somajo;Mood Melodies;James Njie;Thomas Troelsen;Kristoffer Haugan;Edvard Normann;Anders Froen;Gunnar Greve;Yann Bargain;Victor Verpillat;Fredrik Borch Olsen.
2 Sia

I'm Still Here flac

Sia. 2018. Writer: Sia.
3 Cardi B

Taki Taki flac

Cardi B. 2018. Writer: Bava;Juan Vasquez;Vicente Saavedra;Jordan Thorpe;DJ Snake;Ozuna;Cardi B;Selena Gomez.
4 Little Mix

Woman Like Me flac

Little Mix. 2018. Writer: Nicki Minaj;Steve Mac;Ed Sheeran;Jess Glynne.
5 Halsey

Without Me flac

Halsey. 2018. Writer: Halsey;Delacey;Louis Bell;Amy Allen;Justin Timberlake;Timbaland;Scott Storch.
6 Lady Gaga

I'll Never Love Again flac

Lady Gaga. 2018. Writer: Benjamin Rice;Lady Gaga.
7 Bradley Cooper

Shallow flac

Bradley Cooper. 2018. Writer: Andrew Wyatt;Anthony Rossomando;Mark Ronson;Lady Gaga.
8 Bradley Cooper

Always Remember Us This Way flac

Bradley Cooper. 2018. Writer: Lady Gaga;Dave Cobb.
9 Kelsea Ballerini

This Feeling flac

Kelsea Ballerini. 2018. Writer: Andrew Taggart;Alex Pall;Emily Warren.
10 Mako

Rise flac

Mako. 2018. Writer: Riot Music Team;Mako;Justin Tranter.
11 Dewain Whitmore

Burn Out flac

Dewain Whitmore. 2018. Writer: Dewain Whitmore;Ilsey Juber;Emilio Behr;Martijn Garritsen.
12 Avril Lavigne

Head Above Water flac

Avril Lavigne. 2018. Writer: Stephan Moccio;Travis Clark;Avril Lavigne.
13 Khalid

Better flac

Khalid. 2018. Writer: Charlie Handsome;Jamil Chammas;Denis Kosiak;Tor Erik Hermansen;Mikkel Stoleer Eriksen;Khalid.
14 Lady Gaga

Look What I Found flac

Lady Gaga. 2018. Writer: DJ White Shadow;Nick Monson;Mark Nilan Jr;Lady Gaga.
15 Deep Chills

Run Free flac

Deep Chills. 2018.
16 Dynoro

In My Mind flac

Dynoro. 2018. Writer: Georgi Kay;Feenixpawl;Ivan Gough.
17 Charli XCX

1999 flac

Charli XCX. 2018. Writer: Charli XCX;Troye Sivan;Leland;Oscar Holter;Noonie Bao.
18 NCT 127

Regular (English Version) flac

NCT 127. 2018.
19 Lukas Graham

Love Someone flac

Lukas Graham. 2018. Writer: Don Stefano;Morten "Rissi" Ristorp;Morten "Pilo" Pilegaard;Jaramye Daniels;James Alan;David LaBrel;Lukas Forchhammer Graham.
20 Rita Ora

Let You Love Me flac

Rita Ora. 2018. Writer: Rita Ora.

Language

Popular Tags