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(If the title is unclear, I'm looking at infinite cartesian product of $\mathbb{R}$ indexed by $\mathbb{R}$.)

I thought that I had reasoned this rather well, as follows:

$\mathbb{R}^\mathbb{R} = \{f\mid f:\mathbb{R}\rightarrow\mathbb{R}\}$. Note that this includes functions whose range is not $\mathbb{R}$, $\mathbb{R}$ here is just the codomain. Now consider an element $\textbf{x} \in \prod_{\mathbb{R}}\mathbb{R}$. This is an "uncountably infinite sequence," so to speak, so it does represent a function, where the index indicates the $x$ coordinate and the value at that index indicates the $y$ coordinate. (Consider a constant $x_i = 1 \;\forall\; i\in\mathbb{R}$, a straight line.) But this does not account for functions with vertical asymptotes, or functions which are not surjective (e.g. $f(x) = x^2$), so we conclude that $\prod_{\mathbb{R}}\mathbb{R} \subsetneq \mathbb{R}^\mathbb{R}$.

If we union in $\{+\infty, -\infty\}$, then we can get functions with vertical asymptotes, and if we allow coordinates $x_i = \emptyset$, then we get surjective functions as well. (If that even makes sense semantically).

This seems to make sense to me, and I find it a satisfying answer. However, I was reading this wikipedia article which states that:

An uncountable product of metric spaces need not be metrizable. For example, $\mathbb{R}^\mathbb{R}$ is not first-countable and thus isn't metrizable.

Which implicitly states that $\prod_{\mathbb{R}}\mathbb{R} = \mathbb{R}^\mathbb{R}$. So what am I missing? I feel like it must be something rather obvious.

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Why would $\prod_\mathbb R\mathbb R$ be metrizable? –  Asaf Karagila Dec 11 '11 at 20:15
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"But this does not account for functions with vertical asymptotes, or functions which are not surjective." Why? (Neither construction accounts for functions with vertical asymptotes, and both account for functions which are not surjective.) –  Qiaochu Yuan Dec 11 '11 at 20:15
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It seems the problem may be in your definition of uncountable products. By definition, $\prod_\mathbb{R}\mathbb{R}=\mathbb{R}^\mathbb{R}$. –  M Turgeon Dec 11 '11 at 20:18
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You do understand that $X^Y$ means "all functions with domain $Y$ and image contained in $X$", right? It does not mean "all function with domain contained in $Y$ and image $X$", which seems to be what you are imagining in your argument. –  Arturo Magidin Dec 11 '11 at 20:22
    
Something Qiaochu alluded to: functions with vertical asymptotes are not functions on $\mathbb{R}$. The domain is strictly smaller. –  M Turgeon Dec 11 '11 at 20:22
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1 Answer 1

up vote 7 down vote accepted

Your argument about the product starts well, but then it goes off the rails. Why do you say that "it doesn't account for functions that are not surjective"? Or "functions with vertical asymptotes"? It certainly doesn't include functions that are not defined on all of $\mathbb{R}$, but then neither does $\mathbb{R}^{\mathbb{R}}$. As for surjectivity, nothing in the definition implies surjectivity: the "constant tuple" that is $1$ everywhere is an element of $\prod \mathbb{R}$; what makes you think this is surjective?

Remember the very definition of a direct product: if $\{X_i\}_{i\in I}$ is a family of sets, then $$\prod_{i\in I}X_i = \Bigl\{ f\colon I\to \cup X_i \,\Bigm|\, f(i)\in X_i\text{ for each }i\in I\Bigr\}.$$ And if $X$ and $Y$ are sets, then $$X^Y = \{f\colon Y\to X\}$$ that is, all functions with domain all of $Y$ and with $f(y)\in X$ for all $y\in Y$.

So $$\prod_{r\in\mathbb{R}}\mathbb{R} = \Bigl\{ f\colon \mathbb{R}\to\mathbb{R}\,\Bigm|\, f(i)\in\mathbb{R}\text{ for each }i\in\mathbb{R}\Bigr\} = \mathbb{R}^{\mathbb{R}}.$$

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What is it that he should "also remember?" –  analysisj Dec 11 '11 at 20:26
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"Also, Arturo, remember to proof-read and delete partial sentences that you began and then decided would be better elsewhere..." –  Arturo Magidin Dec 11 '11 at 20:32
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