How can I show that a family of decreasing sequences is a closed subset of $\mathbb{R}^\mathbb{N}$ relative to product topology? I am considering taking complements and showing that the complement is open, or showing that the closure of the set is itself. However it is not clear to me what closed subsets of $\mathbb{R}^\mathbb{N}$ relative to product topology look like.
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In general, closed sets in the product topology can be a bit weird. (For example, the Julia sets are closed when viewed in $\mathbb{R}^2$!) – Patrick Stevens Nov 27 '16 at 07:01
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HINT: You don't need to know what closed subsets of $\Bbb R^{\Bbb N}$ look like: you just have to show that the complement of the set of decreasing sequences is open. To do this, you need only show that each sequence that isn't decreasing has an open nbhd in the product that contains no decreasing sequence. If $x=\langle x_n:n\in\Bbb N\rangle$, there are $k,\ell\in\Bbb N$ such that $k<\ell$, but $x_k>x_\ell$. Use $k$ and $\ell$ to define an open nbhd of $x$ that contains no decreasing sequence.
Brian M. Scott
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Sorry but I am still struggling with the subject. If I am to define a neighborhood in $\mathbb{R}^\mathbb{N}$, i need to use a metric on $\mathbb{R}^\mathbb{N}$. Is there a default metric for this space? A previous problem had the metric $d(x,y) = \sum_{n=1}^\infty 2^{-n} \frac{x(n)-y(n}{1+|x(n)-y(n)|}$ for $\mathbb{R}^\mathbb{N}$, should I use that metric? – rush0811 Nov 27 '16 at 08:57
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@rush0811: The space is metrizable, and that is one of the compatible metrics, but using a metric is definitely doing it the hard way. It's much easier to use the definition of the product topology directly. A basic open set of the form $\prod_{n\in\Bbb N}U_n$ with $U_n=\Bbb R$ for each $n$ except $k$ and$\ell$ will work if you choose $U_k$ and $U_\ell$ properly. – Brian M. Scott Nov 27 '16 at 09:32