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I'm doing this exercise in Munkres book and got no clue about the solution. Hope some one can help me solve this.

Show that the product space $R^{I}$, where $I = [0,1]$, has a countable dense subset. If $J$ has cardinality greater than $\mathscr{P}(Z_{+})$, then the product space $R^{J}$ does not have a countable dense subset

With this problem, I even can't imagine the space $R^{I}$. So if some one can talk something about this space, I really appreciate it (as I know, this space is rather important, right?)

Thanks so much

Martin Sleziak
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le duc quang
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  • Actually, $\mathbb{R}^I$ is the space of all functions $I\to \mathbb{R}$ with the topology of pointwise convergence. I don't know if this can help you, though. – Giuseppe Negro Sep 09 '13 at 17:29
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    You might find the answers to this question useful: http://math.stackexchange.com/questions/420384/the-product-space-mathbbri-where-i-denote-0-1-has-a-countable-dense-s?rq=1 – BaronVT Sep 09 '13 at 17:50
  • There is Hewitt—Marczewski—Pondiczery theorem which says $Π_{i < 2^κ} X_i$ has density $≤ κ$ if all $X_i$ have desity $≤ κ$. $\mathbb{R}^I$ is product of continuum separable spaces hence is separable. On the other hand the limit of $2^κ$ spaces in HMP theorem is maximal so product of more than continuum separable spaces is needs not to be separable. – user87690 Sep 09 '13 at 21:26

2 Answers2

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Note that the structure of $I$ or $J$ is irrelevant for topology of $\mathbb{R}^I$ or $\mathbb{R}^J$. Only the cardinality matters. So we want to show that product of continuum copies of $\mathbb{R}$ is separable but product of more than continuum copies of $\mathbb{R}$ is not separable. Actually the following holds: Product of $≤ 2^κ$ spaces of density $≤ κ$ has density $≤ κ$ (for $κ$ infinite cardinal). This is called Hewitt–Marczewski—Pondiczery theorem (e. g. Engelking 2.3.15). On the other hand if all spaces forming a product contain two disjoint open subsets (e. g. at least 2 point Hausdorff spaces) then there is injection from index set to powerset of a dense set so product of more than $2^κ$ such spaces cannot have density $≤ κ$. For proofs and more details see http://dantopology.wordpress.com/2009/11/06/product-of-separable-spaces/.

user87690
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In my answer, I will prioritize intuition over style. (The proof will still be logically rigorous. But you will read more plain English and fewer mathematical expressions.)

For the first part, I will construct a dense subset of $\mathbb{R}^I$.

First, $\mathbb{R}^I$ is conceptually similar to $\mathbb{R}^\omega$. It's just that the index set $I$ is uncountable. We are still dealing with product topology here.

Intuitively, each element $p$ in $\mathbb{R}^I$ can be interpreted as a function $f: [0,1] \rightarrow \mathbb{R}$. A function $f$ with $f(x) = y$, e.g., means that the $x$-th component of the corresponding $p$ has value $y$.

Now we can define the following sets of functions $[0,1] \rightarrow \mathbb{R}$.

  • $D_1$ is the set of all functions on [0,1] that have a constant rational value on [0,1];
  • $D_2$ is the set of all functions on [0,1] that have a constant rational value on [0, 1/2) and another constant rational value on [1/2, 1];
  • $D_3$ is the set of all functions on [0,1] that have a constant rational value on [0,1/4), another constant rational value on [1/4, 2/4), another constant rational value on [2/4, 3/4), and another constant rational value on [3/4, 1].
  • etc.

It is obvious that each $D_i$ is countable. So $D \equiv \cup_i D_i$ is countable. It is not hard to prove that $D$ is a dense subset of $\mathbb{R}^I$. Remember, in the product topology, a basis has the form $\Pi_x U_x$, where $U_x$ is different to $\mathbb{R}$ only for a finite number of $x$'s.

For the second part, follow the hint in Munkres.

Hint: If $D$ is dense in $\mathbb{R}^J$, define $f : J \rightarrow \mathscr{P}(D)$ by the equation $f(\alpha) = D \cap \pi_\alpha^{−1} ((a, b))$, where $(a, b)$ is a fixed interval in $\mathbb{R}$.

Now again, remember each element in $\mathbb{R}^J$ corresponds to a function $\phi: J \rightarrow \mathbb{R}$.

$\pi_\alpha^{−1} ((a, b))$, in plain English, is simply the set of all functions $\phi: J \rightarrow \mathbb{R}$ such that $\phi(\alpha) \in (a, b)$. Since $D$ is dense, $f(\alpha) = D \cap \pi_\alpha^{−1} ((a, b))$ is not empty. Furthermore, it is not hard to show that if $\alpha \neq \beta$, $f(\alpha) \neq f(\beta)$.

All this means that $f$ is injective. Therefore the cardinality of $\mathscr{P}(D)$ is greater than or equal to the cardinality $J$, which is strictly greater than $\mathscr{P}(\mathbb{Z}_+)$ by assumption. Ergo, the cardinality of $D$ is strictly greater than that of $\mathbb{Z}_+$, meaning that $D$ is uncountable.

Petra Axolotl
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