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On page 12 of Luenberger's Optimization by Vector Space Methods, he claims that the set of bounded real sequences is vector space with addition defined component-wise.

Ok, but what is the underlying field? $\mathbb{R}$? But then for a given $K$, you can always find a large enough $M(K)\in \mathbb{R}$ such that $M(K)\left(x_i\right)_{i\in \mathbb{N}}$ becomes unbounded ($\exists i\in\mathbb{N}$ such that $\lvert M(K)x_i\rvert > K$) and so is not closed under scalar multiplication.

Thinking naively, if it is not possible to pick an arbitrarily large scalar, then it should also not be possible to ever construct an unbounded real sequence (as the components of the sequence and the field are the same set, $\mathbb{R}$).

As is probably evident, my knowledge of analysis is rudimentary.

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I think you're confused about "the set of bounded real sequences". It doesn't mean the set of real sequences bounded by some given bound $K$, but the set of all real sequences bounded by some bound, which may be different for each sequence.

That is, we're not talking about the set

$$S_K = \{(x_i)_{i \in \mathbb N}\ |\ \forall i \in \mathbb N: -K \le x_i \le K\},$$

for some given $K$, but about the set

$$S = \{(x_i)_{i \in \mathbb N}\ |\ \exists K \in \mathbb R: \forall i \in \mathbb N: -K \le x_i \le K\} = \bigcup_{K \in \mathbb R} S_K.$$

It's easy to see that $S$ is closed under addition and scalar multiplication. In particular, note that a sequence belongs to $S$ if and only if it belongs to $S_K$ for some $K \in \mathbb R$.

Thus, if $(x_i)_{i \in \mathbb N}$ belongs to $S$, it belongs to $S_K$ for some $K \in \mathbb R$. Therefore, for any $a \in \mathbb R$, $a(x_i)_{i \in \mathbb N}$ belongs to $S_J$, where $J = |a|K$, and thus to $S$.

Similarly, if $(x_i)_{i \in \mathbb N} \in S_K$ and $(y_i)_{i \in \mathbb N} \in S_J$, then $(x_i)_{i \in \mathbb N} + (y_i)_{i \in \mathbb N} \in S_{K+J} \subset S$.

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    Thanks Ilmari. Ok, but then I can have $K(\left(x_i\right))$ as the bound for an individual sequence, and choose arbitrarily large $M(K(\left(x_i\right)))$ that makes that sequence unbounded. I think my confusion is that I am tending to think of $\mathbb{R}$ as $\bar{\mathbb{R}}$, but without this, I cannot see how unbounded sequences can result (that is by picking components from $\mathbb{R}$).2011-08-28
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    @Praetoria: I'm not quite sure what you mean by $K((x_i))$ above, but I edited my answer to include an explicit proof that $S$ is closed under addition and scalar multiplication. I hope it clarifies things.2011-08-28
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    It does to some extent. $K(\left(x_i\right))$ simply denotes the sequence dependent finite bound - same as in your definition. I am just wondering why it cannot be that the $a$ cannot be picked large enough (depending on the sequence dependent bound) so as to make the sequence unbounded - given that we have access to the entire real line. For this, I understand that you cannot make a finite number ($K$) non-finite by multiplying by an element of $\mathbb{R}$, but by that same token, I do not understand how you can get an unbounded sequence when picking all components from $\mathbb{R}$.2011-08-28
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    I think my last comment shows my ignorance of analysis - I suppose unbounded sequences are an asymptotic phenomena - and no _one_ element is "$\infty$".2011-08-28
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    @Praetoria: Yes. A simple example of an unbounded sequence is $(x_i)_{i \in \mathbb N}$, $x_i = i$. You should find it easy to verify that all the elements of the sequence are finite, but that the sequence surpasses every finite bound.2011-08-28