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Alessandro Codenotti
Alessandro Codenotti
10:35 AM
I'm having some troubles following a proof of theorem 19.5 from Oxtoby's "Measure and Category"
Let $X$ be a set of cardinality $\aleph_1$ and let $K$ be a class of elements of $X$ with the following properties:
a) $K$ is a $\sigma$-ideal
b) the union of $K$ is $X$
c) $K$ has a subclass $G$ of cardinality $\le\aleph_1$ with the property that every member of $K$ is contained in some member of $G$
d) the complement of each member of $K$ contains a set of cardinality $\aleph_1$ that belongs to $K$
The proof begins by noting that $A=\{\alpha:0\le\alpha<\omega_1\}$ has cardinality $\aleph_1$ so we can pick an onto map $A\to G$ sending $\alpha\mapsto G_\alpha$. Then he defines $H_\alpha=\bigcup\limits_{\beta\le\alpha}G_\beta$ and $K_\alpha=H_\alpha\setminus(\bigcup\limits_{\beta<\alpha} H_\beta)$
first doubt: Isn't $K_\alpha$ the same as $G_\alpha\setminus(\bigcup\limits_{\beta<\alpha}G_\beta)$?
anyway he then defines $B=\{\alpha\in A:K_\alpha \text{ is uncountable}\}$ and from properties (a),(c) and (d) we obtain that $B$ has no upper bound in $A$. I don't see how to obtain that from the quoted properties
any idea? For reference this is a step in the proof of the Sierpinski-Erdös duality theorem, if you know other sources discussing it that'd be also appreciated
 
 
4 hours later…
Martin Sleziak
Martin Sleziak
2:51 PM
@AlessandroCodenotti Not necessarily. That would bean that $G_\beta\subseteq G_\alpha$ for each $\beta<\alpha$.
@AlessandroCodenotti Suppose that $\gamma$ is an upper bound. We know that $U=\bigcup\limits_{\beta<\gamma}H_\beta$ is in the $\sigma$-ideal.
From d) we get that $X\setminus U$ belongs to the $\sigma$-ideal.
And from c) we have that $X\setminus U\subseteq G_\alpha$ for some $\alpha<\omega_1$.
We also have $X\setminus U\subseteq H_\alpha$ and $\alpha>\gamma$.
Sorry, the above is wrong. Let me start once again. I have applied (d) incorrectly.
From (d) we have that $X\setminus U$ contains an uncountable set from $K$.
I.e., $X\setminus U\supseteq V$, where $|V|=\aleph_1$ and $V$ is from the $\sigma$-ideal.
We also have $V\subseteq G_\alpha$ for some $\alpha$ and it is clear that $\alpha>\gamma$.
From $V\subseteq H_\alpha$ and $V\subseteq X\setminus U$ we should be able to derive that $$V\subseteq H_\alpha \setminus \left(\bigcup_{\beta<\gamma} H_\beta\right).$$
And also something like $$V\subseteq \bigcup_{\gamma\le\beta<\alpha} K_\beta.$$
To derive this, we simply take for each element $v\in V$ the smallest $\beta$ such that $v\in H_\beta$ and we notice that this means $v\in K_\beta$.
Thus $V$ is a subset of a union of countably many countable sets, which is a contradiction.
I would not be surprised if my indices are slightly of in some place (maybe somewhere I need $\le$ instead of $<$ or vice versa). But I guess that roughly this argument should work.
 

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