Let $\mathbb{R}^\mathbb{N}$ be the $\mathbb{R}$-vector space of all real sequences and for $(x_n)_{n\in \mathbb{N}}\in \mathbb{R}^\mathbb{N}$ let

$l^\infty :=${$x=(x_n)_{n\in \mathbb{N}}\in \mathbb{R}^\mathbb{N}|\|x\|_\infty < \infty$}, where as $\|x\|_\infty :=\sup_{n\in \mathbb{N}}|x_n|$.

Show that:

a) $l^\infty$ is an $\mathbb{R}$-vector space.

b) $\|\cdot\|_\infty$ defines a norm on $l^\infty$.

c)$ \bar{U}_1(0)\subseteq l^\infty$ is not compact.

d) Why does c) not contradict the Bolzano-Weierstrass theorem?

I'm learning for my upcoming exam and I'm dealing with metric spaces, sequences and such right now. This is one exercise I found in my textbook and am having quite troubles here.

My approaches so far:

a): I'm not sure what to do here, isn't it already an $\mathbb{R}$-vector space since $(x_n)_{n\in \mathbb{N}}\in \mathbb{R}^\mathbb{N}$?

b) Again, $\|\cdot\|_\infty$ is in the condition for $l^\infty$. Not sure how to show that mathematically. I may need to read further into all kinds of norms. Also had troubles while doing other exercises for metric spaces concerning all kinds of norms.

c) My first approach here was to use the definition of compactness. It's that every sequence has a subsequence which converges to a point in $M$. Meaning that I need to find a subsequence that doesn't converge, right? But I don't know how to construct one.

d) Again, I read about the Bolzano-Weierstrass theorem, which states that every bounded sequence has a converging subsequence, right? Meaning, that, so that it's not a contradiction the sequence must be unbounded. But how can I prove that?

Sorry for my lack of work shown in this one here, but I'm seriously struggling with metric spaces in general.


a) $l^\infty$ is defined as a subset of $\mathbb{R}^{\mathbb{N}}$, namely the subset of bounded sequences. Your task is to verify that this subset is a linear subspace. So it must be non-empty, and closed under addition and scalar multiplication. These are routine verifications.

b) You need to verify the norm properties, - $\lVert x\rVert_\infty \geqslant 0$ for all $x\in l^\infty$ and $\lvert x\rVert_\infty = 0 \iff x = 0$, - $\lVert c\cdot x\rVert_\infty = \lvert c\rvert\cdot \lVert x\rVert_\infty$ for all $c\in \mathbb{R}$ and $x\in l^\infty$, and - $\lVert x+y\rVert_\infty \leqslant \lVert x\rVert_\infty + \lVert y\rVert_\infty$ for all $x,y\in l^\infty$.

c) What you intend to use is the definition of sequential compactness. For metric spaces, that is equivalent to compactness, but for general topological spaces, these are different properties. Maybe you have not yet learned about topological spaces, but it's good to be aware that there is in general a difference. Your idea is good, that's one of the easiest ways to show that the closed unit ball is not compact. It remains to find a sequence in the closed unit ball that has no Cauchy subsequence. Can you find a family of very simple elements of $\overline{U_1(0)}$ such that the distance between any two distinct members of the family is $1$? If your elements are simple enough, you will have no problem giving a sequence with the desired properties.

d) $\overline{U_1(0)}$ is a bounded set, so every sequence in $\overline{U_1(0)}$ is bounded. If the Bolzano-Weierstraß theorem did actually state what you wrote, point c) would contradict the theorem. But there is a further restriction in the premises of the theorem. That condition says that the theorem does not apply to $l^\infty$.

  • $\begingroup$ For a) is it not also wise to verify that there exist identity elements for addition and scalar multiplication in the candidate vector space and underlying field, respectively (clearly the field has an identity, but that the field identity works as an identity for the defined form of scalar multiplication is not always a given)? $\endgroup$ – Todd Wilcox Jun 16 '15 at 13:21
  • $\begingroup$ @ToddWilcox When you are given a nonempty subset of a known vector space and attempt to equip it with the vector space operations, the closure properties are all that must be checked. The algebraic properties are inherited from the larger space. $\endgroup$ – Ian Jun 16 '15 at 13:27
  • $\begingroup$ @ToddWilcox The neutral element for addition follows from non-emptiness, since $0\cdot x = 0$ for any $x$. Of course usually the simplest way to show non-emptiness of a subset $S$ is to show $0\in S$, so then the point is moot. For vector spaces, we demand the unitarity of scalar multiplication ($1\cdot x = x$ for all $x\in V$) in the definition. If we want to show that some subset is a linear subspace of a vector space, that, and the associativity and distributivity relations hold automatically when addition and scalar multiplication don't lead out of the subset. $\endgroup$ – Daniel Fischer Jun 16 '15 at 13:27
  • $\begingroup$ It's not necessary to show that the additive identity of $\mathbb{R^N}$ is bounded and therefore an element of $l^{\infty}$? I guess it would be the zero sequence and therefore trivially bounded and convergent. I'm the kind of proof writer who would put that sentence in anyway. $\endgroup$ – Todd Wilcox Jun 16 '15 at 13:37
  • 1
    $\begingroup$ @ToddWilcox On the one hand, $0\cdot v = 0$, so closedness under scalar multiplication plus non-emptiness suffices to have the additive identity. On the other, to have inverses, note $(-1)\cdot v = (-v)$. $\endgroup$ – Daniel Fischer Jun 16 '15 at 13:49

Hint for c: by applying Bolzano-Weierstrass over and over again, $\overline{U}_1(0)$ is compact in the pointwise sense. In other words if $\left \{ \left \{ x^{(n)}_k \right \}_{k=1}^\infty \right \}_{n=1}^\infty$ is a bounded sequence in $l^\infty$, then $a_n = x^{(n)}_k$ has a convergent subsequence for each $k$. So one way to solve the problem is to find a bounded sequence which converges pointwise but does not converge in the sense of the $l^\infty$ norm. One way to do this is to find a sequence whose norm is constant which nevertheless converges pointwise to zero. Can you come up with one?


I'd like to adress only a), since im in a hurry. Maybe ill edit later tonight!

@ a) Not every set contained in a vector space is again a vector space. For example the set of integers $\mathbb{Z}$ is clearly contained in $\mathbb{R}$, howevever it is not a $\mathbb{R}$-vector space since it is not closed under scalar multiplication for i.e.: $\lambda=\frac{1}{2}$ we get$\lambda \mathbb{Z} \notin \mathbb{Z}$.

In the end you have to show that $l^{\infty}$ fullfills the VS axioms (ie.: is closed under scalar multiplication and addition).


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