Cfrgtkky

If $(A,preccurlyeq)$ is a linear ordering such that $|{yin Amid ypreccurlyeq x}| le aleph_gamma$ for all $xin A$, then $|A|lealeph_gamma$.

Does my attempt look fine or contain logical flaws/gaps? Any suggestion is greatly appreciated. Thank you for your help!

My attempt:

Lemma: If $|S| le aleph_gamma$ and, for all $Ain S$, $|A|le aleph_gamma$, then $|bigcup S|le aleph_gamma$.

By Well-Ordering Principle, there is a well-ordering $preccurlyeq'$ on $A$. Let $V$ be the class of all sets and $rm Ord$ be the class of all ordinals. First, we define function $f:mathcal{P}(A)setminus{emptyset} to A$ by $f(X)=min X$ (with regard to $preccurlyeq'$).

Next, we define function $G:V to V$ by $G(x)=f({ain A mid forall tin {rm ran}(x):t prec a})$ if $x$ is a function and ${ain A mid forall tin {rm ran}(x):t prec a} neq emptyset$, and $G(x)=A$ otherwise. By Transfinite Recursion Theorem, there is a function $F: {rm Ord} to V$ such that $F(alpha)=G(F restriction alpha)$ for all $alpha in {rm Ord}$.

It is not hard to verify (by Hartogs number) that $F(alpha)= A$ for some ordinal $alpha$. Let $lambda=min{alpha in {rm Ord} mid F(alpha)= A}$. Then $F restriction lambda$ is a strictly increasing function with ${rm ran}(F restriction lambda)={rm ran}(F )setminus {A} subseteq A$.

We have $F(lambda)=A$, so ${ain A mid forall tin {rm ran}(F restriction lambda):t prec a} = emptyset$. As $(A,preccurlyeq)$ is linear ordering, it follows that, for every $ain A$, there exists $alpha < lambda$ such that $a preccurlyeq F(alpha)$. Hence $S=bigcup_{alpha<lambda}{a in Amid a preccurlyeq F(alpha)}$.

We have $|{a in Amid a preccurlyeq F(alpha)}| le aleph_gamma$ for all $alpha<lambda$ by assumption. Moreover, $|lambda| =|{rm ran}(F restriction lambda)|=|{rm ran}(F )setminus {A}| le |A|le aleph_gamma$. Hence $|bigcup S|=|bigcup_{alpha<lambda}{a in Amid a preccurlyeq F(alpha)}| le aleph_gamma$ by Lemma. This completes the proof.

Update: Thanks to @Asaf's answer, I figure out that the theorem is incorrectly stated. Instead, it should be

If $(A,preccurlyeq)$ is a linear ordering such that $|{yin Amid ypreccurlyeq x}| < aleph_gamma$ for all $xin A$, then $|A|lealeph_gamma$.

My fix for the wrong part at the end:

We have $|{a in Amid a preccurlyeq F(alpha)}| le aleph_gamma$ for all $alpha<lambda$ by assumption. Moreover, $lambda le aleph_gamma$. If not, $aleph_gamma < lambda$. We have $aleph_gamma=|omega_gamma|=|{rm ran}(F restriction omega_gamma)|$ and ${rm ran}(F restriction omega_gamma) subseteq {y in A mid y le F (omega_gamma)}$. Thus $aleph_gamma le |{y in A mid y le F (omega_gamma)}|$. This contradicts our assumption. Hence $|bigcup S|=|bigcup_{alpha<lambda}{a in Amid a preccurlyeq F(alpha)}| le aleph_gamma$ by Lemma. This completes the proof.

The statement itself is wrong. $omega_1$ itself is an example of a linear ordering where for every $x<omega_1$, $|yinomega_1mid yleq x}|leqaleph_0$, but $|omega_1|=aleph_1>aleph_0$.

You can prove, however, that $aleph_{gamma+1}$ is the upper limit. To do this, first note that there is a well-ordered (under $preccurlyeq$!) subset of $A$ which is not bounded from above. Then prove that this cannot have order type greater than $omega_{gamma+1}$. Now you can apply to your lemma and finish the proof.

Alternatively, you can replace $|{yin Amid ypreccurlyeq x}|leqaleph_gamma$ by a strict inequality to obtain a true statement again.