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to proof $\mathbb{R} ^\mathbb{R}$ is not normal, we use that $\mathbb{N} ^\mathbb{R}$ is closed and not normal. But I have some questions at this point:

  1. why $\mathbb{N} ^\mathbb{R}$ is a closed subset of $\mathbb{R} ^\mathbb{R}$? or equivalently, the complement of is open? how can I write the complement as a union of basic sets?

  2. To see $\mathbb{N} ^\mathbb{R}$ is not normal, we define two closed subsets $H_0$ and $H_1$ ( see $\mathbb{R}^\mathbb{R}$ is not normal).

$H_0=\Big\{\langle n_\xi:\xi<\omega_1\rangle\in X:\forall m\in\Bbb N\setminus\{0\}\big(|\{\xi<\omega_1:n_\xi=m\}|\le 1\big)\Big\}$ , and $H_1=\Big\{\langle n_\xi:\xi<\omega_1\rangle\in X:\forall m\in\Bbb N\setminus\{1\}\big(|\{\xi<\omega_1:n_\xi=m\}|\le 1\big)\Big\}$

But it is not clear for me why these complements are open?

thanks,

Community
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user125176
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1 Answers1

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I will replace $\mathbb R^{\mathbb R}$ by $\mathbb R^J$ for an arbitrary set $J$ to not confuse the different occurrences of $\mathbb R$, and also to stress that there is no topology on $J$.

A basic fact about the product topology is that for each $j\in J$ the map $p_j\colon X^J\to X$, $x=(x_i)_{i\in J}\mapsto x_j$ is continuous.

Regarding the first question, $$\mathbb N^J=\left\{x\in \mathbb R^J\colon \text{$x_j\in\mathbb N$ for all $j$}\right\} =\bigcap_{j\in J}\left\{x\in \mathbb R^J\colon x_j\in\mathbb N\right\} =\bigcap_{j\in J}p_j^{-1}[\mathbb N] , $$ which shows that it is closed, because $\mathbb N$ is a closed subset of $\mathbb R$.

Regarding the second question, $$H_i= \bigcap_{\substack{m\in\mathbb N\setminus\{i\}\\j,k\in\mathbb N, j\ne k}} (p_j,p_k)^{-1}[ (\mathbb N\times\mathbb N)\setminus\{(m,m)\} ], $$ (here $(p_j,p_k)\colon \mathbb N^J\to\mathbb N\times\mathbb N$, $(p_j,p_k)(x)=(x_j,x_k)$), and since $\mathbb N\times\mathbb N$ is discrete this is again closed.

Carsten S
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  • $\mathbb N^J=\left{x\in \mathbb R^J\colon \text{$x_j\in\mathbb N$ for all $j$}\right} =\bigcap_{j\in J}\left{x\in \mathbb R^J\colon x_j\in\mathbb N\right}$ is there any typo error in this line? why you wrote as a intersection? – user125176 Jan 31 '14 at 12:33
  • Because in general ${x\colon \text{$x\in A_j$ for all $j$}}=\bigcap_j A_j$. In this way we translate for all conditions to intersections, which is useful, because intersections of closed sets are closed. – Carsten S Jan 31 '14 at 15:01
  • Now it is clear. Also, regarding my second question $(p_j,p_k)^{-1}(x)=(x_j,x_k)$ is there any typo error? where is $x$? – user125176 Feb 01 '14 at 05:45
  • In general, the set of points has all but countably many constant is closed in $\mathbb{R}^J$? Thanks – user125176 Feb 01 '14 at 05:54
  • There was a typo indeed, I've fixed it. – Carsten S Feb 01 '14 at 08:52