my question is about: show that if $\kappa > 2^\omega$, then the space $2^\kappa$ is not seperable. (kunen, page 86, exercise 4) – ıf there exist a countable dense set $D$ where is the contradiction? since density of $D$ not clear for me in product space $2^\kappa$, I could not say anything about contradiction.
kunen exercise about ccc which is not separable
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2The book gives a hint: "if $D \subset {}^\kappa 2$ is countable, show that there are $\alpha < \beta$ such that $(\forall f \in D)(f(\alpha) = f(\beta))$". This is enough to show that $D$ is not dense, because we can look at the open set all of whose elements are $1$ on $\alpha$ and $0$ on $\beta$, and $D$ does not meet this set. – 2011-06-11
1 Answers
Let's follow Carl's hint. If $D$ is a countable subset of $2^\kappa$, then we may enumerate $D=\{p_n\mid n\in\mathbb{N}\}$, and then associate to each $\alpha<\kappa$ the sequence $\langle p_n(\alpha)\mid n\in\mathbb{N}\rangle$, which is an element of $2^\omega$. Since $\kappa\gt 2^\omega$, it follows by the pigeon-hole principle that there must be two ordinals $\alpha\lt\beta$ giving rise to the same pattern, and thus $p(\alpha)=p(\beta)$ for all $p\in D$. From this, it follows that $D$ is not dense as Carl explains.
It is also interesting to note that the hypothesis $\kappa\gt 2^\omega$ is optimal, as $2^{2^\omega}$ is separable, a fact that is a consequence of David MacIver's answer to this MO question, using the Hewitt-Marczewski-Pondiczery theorem.
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0Thanks for answers. When I try to proof, I take a open set in product space 2^\kappa and I did not think f(\alpha) as a element of the open set. But now it is more clear. If possible, could you – 2011-06-11
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0if possible could you say me any reference about piegon-hole principle. Some statements are very complex. To better understand, any book or paper is exist? – 2011-06-11
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0If $p\in 2^\kappa$, then for any coordinate $\alpha\lt\kappa$, the object $p(\alpha)$ is in $2$, that is, $p(\alpha)$ is either $0$ or $1$. So if you think of $\alpha$ as fixed, and $D=\{p_n\mid n\in\mathbb{N}\}$, then $p_n(\alpha)$ determines a binary $\omega$-sequence as $n$ varies. Since there are only $2^\omega$ many such patterns and $\kappa\gt 2^\omega$, it must be that two different coordinates $\alpha$ and $\beta$ have the same pattern. That, there is some $\alpha$ and $\beta$ such that $p_n(\alpha)=p_n(\beta)$ for all $n$. This violates density. – 2011-06-11
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0THANKS FOR GREAT EXPRESSION. – 2011-06-12
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0Sorry but, I am confused about whether or not Kunen say in the hint – 2011-06-17
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0there exist \alpha,\beta such that \alpha < \beta or the hint is true for all different coordinates. since the cardinality of the sequence
is 2^\omega, \alpha<\beta implies that – 2011-06-17contains more than 2^\omega elements? -
0He was just saying that if $\kappa$ is larger than the continuum, then at least two coordinates must have the same pattern on the elements of $D$, since there are only continuum many such patterns. (After all, if all the patterns were different, then we would have an injective map from $\kappa$ to the continuum, and so it wouldn't really be true that $\kappa$ was larger than the continuum.) – 2011-06-17