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Show that a bounded set \(M\) in \(c_{0}\) is totally bounded if and only if for every \(\varepsilon>0\) there is \(n_{0}\) such that \(\left|x_{n}\right| \leq \varepsilon\) for every \(x \in M\) and \(n \geq n_{0}\). Formulate and prove the analogous result for \(\ell_{p}\) spaces. Hint: Every bounded subset of \(\mathbf{R}^{n_{0}}\) is totally bounded.

Short Answer

Expert verified
Bounded sets are totally bounded using tail properties to cover with finite \( \varepsilon-balls. Applies similarly in \ell_p \ space.

Step by step solution

01

Understanding Total Boundedness

Recall that a set is totally bounded if for every \(\varepsilon > 0\), one can cover the set with a finite number of balls of radius \(\varepsilon\).
02

Property of Bounded Set

Given that \(\left|x_{n}\right| \leq \varepsilon\) for every \(x \in M\) and \(n \geq n_{0}\), means the set elements converges to 0 providing uniform structure for the tail. Analyze this property for total boundedness.
03

Implication for \(c_{0}\)

For any \(\varepsilon > 0\), we can choose a finite number of points within a radius \(\varepsilon\) from any point in \(M\), thus fulfilling the requirement of total boundedness.
04

Analog in &&ell_p\) Spaces

In \( \ell_p \) spaces, consider any bounded set. We must show that for any given \(\varepsilon > 0\), there exists a finite collection of \(\varepsilon\)-balls covering the set.
05

Selection of \( n_{0}\)

For spaces \( \ell_p \), utilize the fact: \( \left( \sum \left| x_{i}\right|^{p} \right)^{1/p}\to \left( \sum_{i=0}^{n_{0}} \left| x_{i}\right|^{p} \right)^{1/p} + 0\), aiding in proving total boundedness using finite radius segments.
06

Final Conclusion

Tie back the small indices (\varepsilon \leq cx_n) to the result of total boundedness in \( \ell_p \). This shows how subsetting indices creates finite sums bounding against \(cx\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Bounded Set
In functional analysis, a set is bounded if there exists a real number that acts as a bound for all the elements in the set. This means that every element in the set is within a certain distance from the origin. If we consider a set of vectors in a normed space, the set is bounded if we can find a number such that the norm of every vector is less than or equal to this number. Essentially, a bounded set does not stretch out to infinity. To illustrate:
  • Imagine a circle within a plane; it is bounded because all its points lie within a finite distance.
  • Conversely, an infinite line is unbounded as points on it can exist at any distance from the origin.
A bounded set ensures that all elements stay within a 'box' of certain size.
c0 Space
The space denoted as \( c_0 \) is a sequence space comprising all sequences of real or complex numbers that converge to 0. To be a bit more formal, if \( x \) is an element of \( c_0 \), then:
  • \( x = (x_1, x_2, ...) \)
  • For any \( \, \epsilon > 0 \), there's an index \( n_0 \) where \( |x_n| < \, \epsilon \) for all \( n \, \) greater than \( n_0 \).
This space is equipped with the sup norm: \[ ||x||_{\infty} = \sup_{n} |x_n| \] One practical example of this is a sequence of numbers that grow smaller and approach 0 as their position in the sequence increases. For example, the sequence \( (1, \, \frac{1}{2}, \, \frac{1}{3}, \, \frac{1}{4}, \, ...) \) belongs to \( c_0 \) because its elements get arbitrarily close to 0.
ℓp Spaces
The \( \ell_p \) spaces are another family of sequence spaces, where \( p \) is a real number greater than or equal to 1. An element in \( \ell_p \) is a sequence of numbers \( x = (x_1, x_2, ...) \) such that the following series is convergent: \[ ||x||_p = \left( \sum_{n=1}^{\infty} |x_n|^p \right)^{1/p} \] Different values of \( p \) give us different spaces. For instance:
  • When \( p = 1 \), we get \( \ell_1 \), the space of absolutely summable sequences.
  • When \( p = 2 \), we get \( \ell_2 \), the space of square-summable sequences, which is important in many areas of mathematics and physics.
In general, \( \ell_p \) spaces allow us to generalize the idea of distance and other geometric properties to more abstract sequence spaces.
Totally Bounded Set
A set in a metric space is called totally bounded if, for any given \( \, \epsilon > 0 \), it is possible to cover the set with a finite number of subsets, each having a diameter smaller than \( \, \epsilon \). This concept ensures that wherever you are in the set, you're not far from some point in a finite covering. One way to picture this is to imagine trying to cover a set of points in a plane with circles (or balls in higher dimensions):
  • If you can achieve this with a finite number of circles of arbitrary small size, then the set is totally bounded.
This concept is crucial because a totally bounded set will always be bounded, but the converse isn't necessarily true. Confirming total boundedness often plays a critical role in proofs and theoretical analysis, especially in Functional Analysis.
Mathematical Proof
A mathematical proof is a logical argument that demonstrates the truth of a given proposition or theorem. The goal is to show that, starting from a set of assumed axioms and using deductive reasoning, the proposition must be true. There are different types of proofs, including direct proofs, indirect proofs, and proofs by contradiction. For example, in the provided exercise:
  • You might start with the definition of total boundedness and use given properties of \( c_0 \) or \( \ell_p \) spaces.
  • You logically deduce that a bounded subset must necessarily be totally bounded under the given conditions.
Key to any proof is ensuring every step follows logically from the previous one and all assumptions and conditions are clear. Mathematical proofs not only establish the truth of a statement but also enhance our understanding of why it's true.

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Most popular questions from this chapter

Let \(M\) be a dense subset (not necessarily countable) of a Banach space \(X\). Show that for every \(x \in X \backslash\\{0\\}\) there are \(x_{k} \in M\) such that \(x=\sum x_{k}\) and \(\left\|x_{k}\right\| \leq \frac{3\|x\|}{2^{k}} .\) Hint: Find \(x_{1} \in M\) such that \(\left\|x-x_{1}\right\| \leq \frac{\|x\|}{2} ;\) then by induction find \(x_{k} \in M\) such that \(\left\|x-\left(x_{1}+\ldots+x_{k-1}\right)-x_{k}\right\| \leq \frac{\|x\|}{2^{k}} .\) Then \(x=\sum x_{k}\) and \(\left\|x_{k}\right\|=\left\|\left(x-\sum_{n=1}^{k-1} x_{n}\right)-\left(x-\sum_{n=1}^{k} x_{n}\right)\right\| \leq \frac{\|x\|}{2^{k-1}}+\frac{\|x\|}{2^{k}}=\frac{3\|x\|}{2^{k}}\)

Show that a Banach space \(X\) is separable if and only if \(S_{X}\) is separable. Hint: If \(\left\\{x_{n}\right\\}\) is dense in \(S_{X}\), consider \(\left\\{r_{k} x_{n}\right\\}_{k, n}\) for some dense sequence \(\left\\{r_{k}\right\\}\) in \(\mathbf{K}\). If \(\mathcal{S}\) is countable and dense in \(X\), consider \(\left\\{\frac{x}{\|x\|} ; x \in \mathcal{S}\right\\}\).

Let \(C\) be a convex set in a normed space \(X\), assume \(\operatorname{Int}(C) \neq \emptyset\) (recall that \(\operatorname{Int}(C)\) denotes the interior of \(C\) ). Show that \(\overline{\operatorname{Int}(C)}=\bar{C}\) and \(\operatorname{Int}(\bar{C})=\operatorname{Int}(C)\) Hint: Use the "cone" argument. There is a point \(x_{0}\) and an open ball \(B_{b}^{\circ}\) such that \(x_{0}+B_{\delta}^{O} \subset C .\) Note that if we take any \(c \in C\), then by the convexity of \(C\), the cone with the vertex \(c\) and the base \(x_{0}+B_{\delta}^{O}\) is a subset of \(C\), and all points in this cone but the vertex \(c\) are its interior points (draw a picture!). Therefore, given \(c \in C\), we can find points from \(\operatorname{Int}(C)\) arbitrarily close to \(c\), showing that \(\bar{C} \subset \overline{\operatorname{Int}(C)}\) Given \(c \in \operatorname{Int}(\bar{C})\), there is \(\varepsilon>0\) such that \(c+\varepsilon B_{X} \subset \bar{C} .\) Let \(d=\) \(c+\varepsilon\left(c-x_{0}\right) .\) Then \(d \in \bar{C}\), so for every \(\nu>0\) there is \(y \in C\) with \(\|d-y\|<\nu\) If \(\nu\) is small (namely \(\left.\nu \leq \frac{\varepsilon \delta}{\sqrt{\delta^{2}+\left\|x_{0}-c\right\|^{2}}}\right), c\) is in the cone with the base \(x_{0}+B_{\delta}^{O}\) and vertex \(y\), so \(c \in \operatorname{Int}(C)\).

Let \(X\) be a normed space, \(M \subset X\) and \(\varepsilon>0 .\) We say that \(A \subset M\) is an \(\varepsilon=n e t\) in \(M\) if for every \(x \in M\) there is \(y \in A\) such that \(\|x-y\|<\varepsilon\). We say that \(A \subset M\) is \(\varepsilon\) -separated in \(M\) if \(\|x-y\| \geq \varepsilon\) for all \(x \neq y \in A\). Clearly, a maximal \(\varepsilon\) -separated subset of \(M\) in the sense of inclusion is an \(\varepsilon\) -net in \(M\). Let \(X\) be an \(n\) -dimensional real normed space. Show that if \(\left\\{x_{j}\right\\}_{j=1}^{N}\) is an \(\varepsilon\) -net in \(B_{X}\), then \(N \geq \varepsilon^{-n} .\) On the other hand, there is an \(\varepsilon\) -net \(\left\\{x_{j}\right\\}_{j=1}^{N}\) for \(B_{X}\) with \(N \leq\left(\frac{2}{c}+1\right)^{n}\). Hint: Fix a linear isomorphism \(T\) of \(X\) onto \(\mathbf{R}^{n}\). For a compact subset \(C\) of \(X\), we define the volume of \(C\) by \(\operatorname{vol}(C)=\lambda(T(C))\), where \(\lambda\) is the Lebesgue measure on \(\mathbf{R}^{n}\). Let \(B(x, r)\) be the closed ball centered at \(x\) with radius \(r\). We have \(B_{X} \subset \bigcup_{j=1}^{N} B\left(x_{j}, \varepsilon\right)\), so $$ \begin{aligned} \operatorname{vol}\left(B_{X}\right) & \leq \operatorname{vol}\left(\bigcup_{j=1}^{N} B\left(x_{j}, \varepsilon\right)\right) \leq \sum_{j=1}^{N} \operatorname{vol}\left(B\left(x_{j}, \varepsilon\right)\right) \\ &=N \cdot \operatorname{vol}\left(\varepsilon B_{X}\right)=N \cdot \varepsilon^{n} \cdot \operatorname{vol}\left(B_{X}\right) \end{aligned} $$ Thus \(N \geq \varepsilon^{-n}\). On the other hand, if \(\left\\{x_{j}\right\\}_{j=1}^{N}\) is a maximal \(\varepsilon\) -separated set in \(B_{X}\), then \(\left\\{x_{j}\right\\}_{j=1}^{N}\) is an \(\varepsilon\) -net in \(B_{X}\). Since \(B\left(x_{j}, \varepsilon / 2\right) \cap B\left(x_{i}, \varepsilon / 2\right)=\emptyset\) if \(i \neq j\) and \(B\left(x_{j}, \varepsilon / 2\right) \subset(1+\varepsilon / 2) B_{X}\), we have $$ \begin{aligned} N \cdot(\varepsilon / 2)^{n} \cdot \operatorname{vol}\left(B_{X}\right) &=\operatorname{vol}\left(\bigcup_{j=1}^{N} B\left(x_{j}, \varepsilon / 2\right)\right) \\ & \leq \operatorname{vol}\left((1+\varepsilon / 2) B_{X}\right)=(1+\varepsilon / 2)^{n} \operatorname{vol}\left(B_{X}\right) . \end{aligned} $$ Thus \(N \leq\left(\frac{1+\varepsilon / 2}{\varepsilon / 2}\right)^{n}\).

Let \(H\) be a Hilbert space. Prove the generalized parallelogram equality: If \(x_{1}, \ldots, x_{n} \in H\), then $$ \sum_{\epsilon_{i}=\pm 1}\left\|\sum_{i=1}^{n} \varepsilon_{i} x_{i}\right\|^{2}=2^{n} \sum_{i=1}^{n}\left\|x_{i}\right\|^{2} $$ Hint: Induction on \(n\).

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