Let $\mathbb{R}^\omega$ be the infinite product space $\mathbb{R} \times \mathbb{R} \times ...$, and let $A$ be the subset of $\mathbb{R}^\omega$ containing points with coordinates $0$ except for a finite amount of coordinates. The following proof contains an error:

The set $A$ can be broken up into subsets of the form $\prod X_\alpha$ where $X_\alpha$ is $\mathbb{R}$ for a finite amount of $\alpha$ and $\{0\}$ otherwise. Of course, $A$ is then the union of all such sets.

Take an arbitrary coordinate $\beta$. We can find a subset $\prod X_\alpha \subset A$ as described before such that $X_\beta = \mathbb{R}$. Since the choice of $\beta$ was arbitrary, we can conclude that in the union of all such subsets, each coordinate maps to $\mathbb{R}$ and therefore $A$ can be expressed as $\prod X_\alpha$ where $X_\alpha = \mathbb{R}$ for all $\alpha$.

Let $x = (1, 1, ...)$. Since each $x_\alpha$ is in $\mathbb{R}$ we can conclude $x \in A$ for the derived definition of $A$. However, $x \notin A$ for the given definition of $A$, a contradiction.

I don't see a reason why the definition of $A$ cannot exist, so there must be something wrong with the proof, but I can't find the problem.


2 Answers 2


The issue is with your claim

$(*)\quad$ and therefore $A$ can be expressed as $\prod X_\alpha$ where $X_\alpha = \mathbb{R}$ for all $\alpha$.

Essentially, all you've shown up until this point is that $A$ isn't equal to any of its "finite-length approximations." But this doesn't imply that $A=\prod_{\alpha\in\omega} \mathbb{R}$ (and so your final paragraph is a non-starter, since $x\not\in A$).

If you try to prove $(*)$, I suspect you'll quickly see the issue.


Noah pointed out the mistake. But let me make an even stronger point:

The union of products need not be a product.

To see this, note for example that $\{\langle x,y\rangle\in\Bbb R^2\mid |x|,|y|<1\}$ is the union of countably many sets of the form $(a,b)\times(c,d)$. But the unit disc itself is not a product of any two subsets of $\Bbb R$.

Even worse, $\Bigr((0,2)\times(0,2)\Bigl)\cup\Bigr((3,5)\times(3,5)\Bigl)$ is a union of two products and it is not a product, since $\langle 1,4\rangle$ is not in the set, and if it were a product of two sets $A$ and $B$, then $1\in A$ since $\langle 1,1\rangle$ is in the set, and $4\in B$ for a similar reason.


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