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To start, suppose that $\lambda$ is an infinite cardinal and suppose that $\alpha, \beta \in \mathbf{H}_\lambda$, where $\mathbf{H}_\lambda = \{x : \left|\operatorname{trcl}(x)\right| < \lambda\}$. I'm wondering, is it true that $\alpha \times \beta \in \mathbf{H}_\lambda$? I feel as though it ought to be, but I'm really not sure how to show it.

Explicitly, the notion of transitive closure I've been working with is defined as $\operatorname{trcl}(x) = \bigcup_{n \in \mathbf{N}} U(x, n)$, where $U(x, 0)= x$ and $U(x, n+1)= \bigcup U(x,n)$. So my thought was that if one could show by induction that $\left|U(\alpha \times \beta, n)\right| < \lambda$ for all $n \in \mathbf{N}$, then we could conclude that $|\operatorname{trcl}(\alpha \times \beta)| < \lambda$, but I'm having a hard time finishing the induction. I was hoping someone could help me sort this out? Any assistance is appreciated.

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Remember the definition of the Cartesian product: $$A\times B=\{\{\{a\},\{a,b\}\}:a\in A,\>b\in B\}.$$

From this, we get $$ \begin{align} \operatorname{trcl}(A\times B)=&(A\times B)\cup\bigcup(A\times B)\cup\bigcup{}^2(A\times B) \cup \operatorname{trcl} \bigcup{}^2(A\times B)\\ =&(A\times B)\cup (\{\{a\}:a\in A\}\cup \{\{a,b\}:a\in A,b\in B\})\cup (A\cup B)\cup \operatorname{trcl}(A\cup B) \end{align} $$ ($\bigcup^2 X$ means $\bigcup\bigcup X$.)

and you can check that each set mentioned above has cardinality $<\lambda$.

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  • $\begingroup$ This is really nice, thanks! $\endgroup$ Commented Dec 16, 2014 at 7:34

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