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Chapter 6: Problem 4

Show that the union of finitely many compact sets is compact.

Short Answer

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Question: Prove that the union of finitely many compact sets is compact. Answer: To prove that the union of finitely many compact sets is compact, we assume an open cover of the union and demonstrate that there exists a finite subcover for the entire union. This is done by finding finite subcovers for each compact set and combining them to create a finite subcover for the entire union. The proof concludes that since we have found a finite subcover for the union, the union of finitely many compact sets is compact.

Step by step solution

01

Assume an open cover of A

Let \(\{U_i\}\) be an open cover of \(A\). That is, each \(U_i\) is an open set, and we have \(A \subseteq \bigcup_{i} U_i\). Since \(A\) is the union of the compact sets, this means \(A_1 \cup A_2 \cup ... \cup A_n \subseteq \bigcup_{i} U_i\).
02

Find finite subcovers for each compact set

Since \(A_1, A_2, ..., A_n\) are compact, for each set there exists a finite subcover of \(\{U_i\}\) that covers the corresponding set. Denote these finite subcovers as \(\{U_{1j}\}, \{U_{2j}\}, ..., \{U_{nj}\}\), respectively. This means: 1. \(A_1 \subseteq \bigcup_{j} U_{1j}\) 2. \(A_2 \subseteq \bigcup_{j} U_{2j}\) 3. ... 4. \(A_n \subseteq \bigcup_{j} U_{nj}\)
03

Combine finite subcovers to cover the union of compact sets

Since \(\{U_{1j}\}, \{U_{2j}\}, ..., \{U_{nj}\}\) are finite subcovers for each compact set, we can combine them to create a finite subcover for the entire union: \(A = A_1 \cup A_2 \cup ... \cup A_n \subseteq \bigcup_{j} U_{1j} \cup \bigcup_{j} U_{2j} \cup ... \cup \bigcup_{j} U_{nj}\)
04

Conclude the proof

Since we have found a finite subcover for the union of compact sets \(A\), we can conclude that \(A\) itself is compact. Therefore, the union of finitely many compact sets is also compact.

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Key Concepts

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

Open Cover
Imagine you have a collection of blankets that are large enough to entirely cover you when you're chilly. In mathematics, we use a similar concept for sets, which is called an open cover. More precisely, an open cover of a set in a topological space is a collection of open sets whose union completely encloses the set in question.

For instance, let's consider a set, akin to how you'd be lying on a field. Now, if we take several transparent sheets, each one representing an open set, and lay them over you so that every part of you is covered鈥攖hat forms what we call an 'open cover'. The beauty is that some sheets may overlap, and you might be covered multiple times, but as long as every part of you is under at least one sheet, we're satisfied that it鈥檚 an open cover.
Finite Subcover
Now let's say we have far too many sheets, and we want to be sustainable鈥攐nly using what's necessary to keep you covered. What if we could select just a few of these sheets that still do the job? This action refers to a finite subcover. It鈥檚 essentially a smaller collection of the open sets from our original 'open cover' that still completely covers our set.

This concept is at the heart of 'compactness'. A set is compact if, whenever it has an open cover, it also possesses a finite subcover that still envelops the entire set. In this context, the compactness property ensures we can always find a 'minimal' cover鈥攎uch like selecting the fewest possible blankets needed to keep warm. In our textbook exercise, this step was crucial for proving that the union of several compact sets maintained the desirable trait of being compact.
Union of Compact Sets
When the winter is harsh, sometimes one blanket isn't enough, so we gather more. Similarly, sets can be bundled together to form a bigger set. The exercise you encountered focuses on the concept of the union of compact sets.

This idea is like taking several 'warm' and compact 'blankets' (compact sets) and tossing them together to form a bigger one. As each compact blanket can be covered by a small number of sheets (finite subcover), when you combine all these separate layers into a bigger blanket, you still end up with a 'warm' bundle that can be covered by a finite combination of your original sheets.

This property is vital in topology and analysis as it underpins the behavior of functions and spaces in more complex scenarios. By showing that the combination of a finite number of compact sets is still compact, we ensure that problems can be 'localized' to manageably small pieces, just like using several small blankets efficiently rather than one massive one.

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Most popular questions from this chapter

Let \(A\) be a compact subset and \(B\) be closed subset of the normed linear space \(X\). If \(A\) and \(B\) are disjoint sets show that $$ \inf \\{\|a-b\|: a \in A, b \in B\\}>0 \text {. } $$

Let \(\left\\{F_{n}\right\\}\) be a decreasing sequence of closed sets in a normed linear space \(X\). If \(\cap_{1}^{\infty} F_{n}\) is contained in an open set \(\mathcal{O}\), show that \(F_{N} \subset \mathcal{O}\) for some \(N\).

Let \(A\) be a compact subset of a normed linear space \(X\). Let \(x\) be a point in \(X\) and \(\lambda\) a real number. Show that the sets \(A_{1}=\\{x+a: a \in A\\}\) and \(A_{2}=\\{\lambda a: a \in A\\}\) are both compact.

Let \(f_{n}(s)=s e^{-n s}, n \geq 1\). Show that \(f_{n} \rightarrow 0\) uniformly on \([0, \infty)\).

This question investigates the Binomial series. (i) Show that \(\sum_{n=0}^{\infty}\left(\begin{array}{c}\alpha \\\ n\end{array}\right) s^{n}\) converges for \(|s|<1 .\) Recall that \(\alpha \in \mathbb{R}\) and $$ \left(\begin{array}{c} \alpha \\ n \end{array}\right)=\frac{\alpha(\alpha-1) \ldots(\alpha-n+1)}{n !} $$ (ii) If \(f(s)=\sum_{n=0}^{\infty}\left(\begin{array}{c}\alpha \\\ n\end{array}\right) s^{n}\), show that \((1+s) f^{\prime}(s)=\alpha f(s)\) where \(|s|<1\) Show that any function satisfying (ii) must be of the form \(f(s)=\) \(c(1+s)^{\alpha}\) for some constant \(c \in \mathbb{R} .\) Use this fact to establish that $$ (1+s)^{\alpha}=\sum_{n=0}^{\infty}\left(\begin{array}{c} \alpha \\ n \end{array}\right) s^{n} \quad \text { for }|s|<1 $$ Here the equality means that the series on the right is uniformly convergent to the function on the left of the equality.
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