I cannot prove the fact in the title. Please help! I am reading the handbook of set theoretic topology. And I found this fact in a proof in the book.
Thank you.
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1For the determination of the used nomenclature: A normal space is one that is $T_4$ and $T_1$ in that handbook? – Daniel Fischer Jul 02 '15 at 09:43
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To be more precise, I am reading the chapter 18 of the book. As mentioned in the introduction, the author follows the definition in Engelking's book. The author also assumes that all spaces are completely regular. – Tahina_RAK Jul 02 '15 at 10:44
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Is it true that $|C(X)|\ge |\tau|$ for a normal space $(X,\tau)$, where $C(X)$ is the set of real-valued continuous functions on $X$? ($|C(X)|\le c$ for separable $X$.) – David Mitra Jul 02 '15 at 11:38
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Hi David, can please tell me why |C(X)|$ \leq$ c for separable X – Tahina_RAK Jul 02 '15 at 14:36
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See the first answer here. – David Mitra Jul 02 '15 at 14:44
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@Jessy: Exactly where in Teodor’s chapter did you find this assertion? – Brian M. Scott Jul 02 '15 at 21:02
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1@BrianM.Scott: please read the answer provide by hot_queen below. – Tahina_RAK Jul 03 '15 at 14:29
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This is false: $X = 2^{2^{\omega}}$ under the usual product topology is a counterexample. It is obviously normal being compact Hausdorff. Checking that $X$ is separable is a fun exercise. Finally, $X$ has more than continuum many points and hence more than continuum many open sets. The correct upper bound is $2^{2^{\omega}}$ and you can prove this by showing that $\{\{x \in X: f(x) \neq 0\} \mid f:X \to \mathbb{R}$ continuous$\}$ is a basis of size at most continuum. The above example shows that this bound is optimal.
Addendum: I looked at chapter 18 and noticed that the proof of Jones' lemma on page 784 contains the pharse "Since $X$ is separable, it has only continuum many different open sets ..." which is incorrect. The correct argument goes as follows: Given $F$ discrete closed in $X$, for each $K \subseteq F$, get an open set $U_K$ such that $K \subseteq U_K, \overline{U_K} \cap (F \setminus K) = \phi$. Let $D$ be countable dense in $X$. Suppose $K, K'$ are distinct subsets of $F$ and say $x \in K \setminus K'$. Let $x \in V \subseteq U_{K}$ where $V$ is open and disjoint with $U_{K'}$. Then $D \cap V \subseteq U_K \backslash U_{K'}$. Hence the map $K \mapsto U_K \cap D$ is injective. So $2^{|F|} \leq 2^{|D|} = 2^{\aleph_0}$.
hot_queen
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1Right, $X=\beta \mathbb{N}$ can serve as another counterexample as it is separable by the very definition. – Tomasz Kania Jul 02 '15 at 22:04
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@ hot_queen: thank you for your answer. I am still new in this field so, could you please tell me what is the countable dense set in $X= 2^{2^{\omega}}$? – Tahina_RAK Jul 03 '15 at 14:59
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@ Tomek kania: same question as above, please provide me the countable dense set in your given space – Tahina_RAK Jul 03 '15 at 15:01
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For each $n < \omega$ and $A \subseteq 2^n$, define $f_A \in 2^{2^{\omega}}$ by $f_{A}(x) = 1$ iff $x$ is a branch through a node in $A$. Try to show that ${f_A : n < \omega, A \subseteq 2^n}$ is dense in $2^{2^{\omega}}$. – hot_queen Jul 03 '15 at 15:06
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