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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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    @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