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Could anyone explain to me why the set of all bounded sequences of real numbers is open in the box topology? Can we say the same for the set of all unbounded sequences?

Martin Sleziak
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user391120
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2 Answers2

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If a sequence $(x_n)$ is bounded, so is every sequence in its box open neighbourhood $U = \prod_n (x_n -1 ,x_n +1)$. So bounded sequences form an open subset.

If that sequence is unbounded, so is every sequence in $U$ as well, making the unbounded sequences open as well.

So the set of bounded sequences is both open and closed. This shows that the infinite box product of connected spaces need not be connected.

Henno Brandsma
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  • :How do we conclude that a bounded sequence has a neighbourhood like this one? – user391120 Dec 08 '16 at 13:36
  • @user391120 I have defined a basic box open set, haven't I? – Henno Brandsma Dec 08 '16 at 13:38
  • Yes you have. My question is if we can do the same for any sequence or it is relative with a dounded or unbounded sequence only? – user391120 Dec 08 '16 at 13:43
  • @user391120 we can define it for any sequence. If the $x_n$ are bounded by, say, $m$, all sequences in this neighbourhood are bounded by $m+1$. If the sequence is unbounded, then so are all sequences in that neighbourhood (why?). – Henno Brandsma Dec 08 '16 at 13:47
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Let $U_n = \{(x_i)\mid |x_i| < n \text{ for all }i\in \Bbb N\}$, i.e. all sequences that never exceed $n$. This is an open set, because it's an element in the standard box topology basis. The set of all bounded sequences is the union of all these open $U_n$'s.

Similarily, an unbounded sequence is one that is not bounded, so the set of unbounded sequences is the complement of the set of bounded sequences. Therefore the set of unbounded sequences is closed in the box topology.

One would think we could do something similar with unbounded sequences, i.e. define the open $V_n = \{(x_i)\mid |x_i| > n \text{ for at least one } i\in \Bbb N\}$. This does not work because in this case we want the intersection of all the $V_n$, and an infinite intersection of open sets is not necessarily open. In fact, if you use $\geq$ in the definition of $V_n$ instead, they become closed (they're the complements of the $U_n$), and you get that the set of unbounded sequences is an intersection of closed sets, and therefore closed.

Arthur
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  • Could you please write some elemements of set Un because i cannot understand why is this an element in the box topology basis? – user391120 Dec 08 '16 at 13:31
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    $U_1$ is the basis element $(-1,1)\times (-1,1)\times \cdots$, so, for instance, $(1/2, 1/3,1/4,1/5,\ldots)$ and $(1/2,-3/4,7/8,-15/16,\ldots)$ are in there. So is a lot of other sequences, just as long as (the absolute value of) each element is strictly less than $1$. – Arthur Dec 08 '16 at 13:52