Question

Let \(\sum x_{i}\) be unconditionally convergent. Show that: (i) \(S_{1}=\left\\{\sum_{i=1}^{\infty} \varepsilon_{i} x_{i} ; \varepsilon_{i}=\pm 1\right\\}\) is a compact set in \(X\), (ii) \(S_{2}=\left\\{\sum_{i=1}^{n} \varepsilon_{i} x_{i} ; n \in \mathbf{N}, \varepsilon_{i}=\pm 1\right\\}\) is a relatively compact set in \(X\). Hint: (i): The space \(\\{-1,1\\}^{\mathrm{N}}\) is compact in the pointwise topology. We claim that the map \(\left\\{\varepsilon_{i}\right\\} \mapsto \sum \varepsilon_{i} x_{i}\) is continuous. Indeed, from the unconditional Cauchy condition we have that if \(\sigma\) is a finite set with \(\min (\sigma)\) sufficiently large, then \(\left\|\sum_{i \in \sigma} x_{i}\right\|<\varepsilon\), which gives that if \(n_{0}\) is sufficiently large, then \(\left\|\sum_{i=n}^{m} \varepsilon_{i} x_{i}\right\|<\varepsilon\) for every possible \(\varepsilon_{i}\). (ii): Use (i) and write \(2 \sum_{i=1}^{n} \varepsilon_{i} x_{i}=\sum_{i=1}^{\infty} \varepsilon_{i} x_{i}+\sum_{i=1}^{\infty} \varepsilon_{i}^{\prime} x_{i}\), where \(\varepsilon_{i}^{\prime}=\varepsilon_{i}\) for \(i=1, \ldots, n\) and \(\varepsilon_{i}^{\prime}=-\varepsilon_{i}\) for \(i=n+1, n+2, \ldots\)

Short answer

Set \(S_{1}\) is compact, and set \(S_{2}\) is relatively compact.

Step-by-step solution

Understand the given sets

We have two sets: \(S_{1}\) and \(S_{2}\). Set \(S_{1}\) represents all possible infinite series formed by taking elements \( x_{i} \) multiplied by \( \varepsilon_{i} \) which are either \( +1 \) or \( -1 \). Set \(S_{2}\) represents finite sums of a similar form up to an index \( n \), with \( n \in \mathbf{N} \).

Compactness of set \(S_{1}\) (Hint usage)

From the hint, we recognize that the space \( \{-1,1\}^\mathbb{N} \) is compact in the pointwise topology. To show that the map \( \{\varepsilon_{i}\} \mapsto \sum \varepsilon_{i} x_{i} \) is continuous, note that from the unconditional convergence given, for any finite set \(\sigma\) where \( \min (\sigma) \) is large enough, we have: \[ \left\|\sum_{i \in \sigma} x_{i} \right\| < \varepsilon \] This implies for \( n_0 \) large enough: \[ \left|\sum_{i=n}^{m} \varepsilon_{i} x_{i} \right| < \varepsilon \] for any choice of \( \varepsilon_{i} \). Thus, the map is continuous, implying that \(S_{1}\) is the continuous image of a compact set, hence compact.

Relative compactness of \(S_{2}\)

Using the result from (i) and analyzing set \( S_{2} \): Any element of \( S_{2} \) can be written using two series from the set \( S_{1} \). Specifically, let: \[ 2 \sum_{i=1}^n \varepsilon_{i} x_{i} = \sum_{i=1}^{\infty} \varepsilon_{i} x_{i} + \sum_{i=1}^{\infty} \varepsilon_{i}^{\prime} x_{i} \] where \ \( \varepsilon_{i}^{\prime} = \varepsilon_{i} \) for \ \( i = 1, \ldots, n \) and \ \( \varepsilon_{i}^{\prime} = -\varepsilon_{i} \) for \ \( i = n+1, n+2, \ldots \). Each \( n \)-term partial sum is a linear combination of two elements from \(S_{1}\), thus falling within a relatively compact set owing to the compactness of \(S_{1}\).

Key concepts

compact set

A compact set in mathematics has a very important property: it is both closed and bounded. This means that every sequence of points within the set has a convergent subsequence that converges to a point still within the set. In context, for set \(S_{1}\), as mentioned in the exercise, we look at the space \(\{-1,1\}^\mathbb{N}\). This space is known to be compact in the pointwise topology.
By using the map \(\{\varepsilon_{i}\} \rightarrow \sum \varepsilon_{i} x_{i}\), we see that it is continuous, given the initial condition of unconditional convergence.
This means if we choose any sequence of points from set \(S_{1}\), its subsequence will converge to a point within \(S_{1}\). Hence, proving \(S_{1}\) is compact since the image of a compact set under a continuous map is also compact.

relative compactness

Relative compactness is about whether a set's closure is compact. To break it down, \(S_{2}\) is assessed based on being within the closure of another compact set from the exercise.
Imagine \(S_{2}\) as consisting of finite sums up to some index. Since each element of \(S_{2}\) can be represented using elements of \(S_{1}\), we look at two infinite series related to \(S_{1}\).
We consider \(2 \sum_{i=1}^{n} \varepsilon_{i} x_{i} = \sum_{i=1}^{\infty} \varepsilon_{i} x_{i} + \sum_{i=1}^{\infty} \varepsilon_{i}^{\prime} x_{i}\).
By separating the first \(n\) terms and their corresponding parts, we ensure every element of \(S_{2}\) stays compact, thus relatively compact. It’s about ensuring the overarching structure (closure of \(S_{2}\)) retains compactness properties.

continuity

Continuity is a fundamental concept where small changes in the input result in small changes in the output. In the exercise, continuity is crucial in understanding the transition between the sequences and their sums.
For the map \(\{\varepsilon_{i}\} \rightarrow \sum \varepsilon_{i} x_{i}\), noting continuity means that for any two points that are close in the input space, their mapped images are also close in the output space.
If \(\{\varepsilon_{i}\}\) changes slightly, its sum \(\sum \varepsilon_{i} x_{i}\) changes slightly as well. This stability is built on the Cauchy condition, ensuring the control over the sums. It is exactly this aspect that ensures the integrity and validity of the sets \(S_{1}\) and \(S_{2}\) as explored in the exercise. This fundamental stability (continuity) thus helps in proving compactness and relative compactness.