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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.

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    The 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

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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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    He 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