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Give examples to show that an infinite intersection of open sets may not be open, and an infinite union of closed sets may not be closed. [Hint: Show that $$ \bigcap_{n=1}^{\infty}\left(-\frac{1}{n}, \frac{1}{n}\right)=\\{0\\} $$ and $$ \left.\bigcup_{n=2}^{\infty}\left[\frac{1}{n}, 1-\frac{1}{n}\right]=(0,1) .\right] $$

Short Answer

Expert verified
The intersection is \(\{0\}\), not open; the union is \((0,1)\), not closed.

Step by step solution

01

Understanding Open Sets and Intersections

An open set on the real line is a set that does not contain its boundary, so intervals like \(-\frac{1}{n}, \frac{1}{n}\) are open. When considering the intersection of an infinite number of such open sets, we aim to find points common to all sets.
02

Solving the Intersection Example

Consider each set \(-\frac{1}{n}, \frac{1}{n}\). As \(n\) increases, \(-\frac{1}{n}\) gets closer to zero from the negative side and \(\frac{1}{n}\) from the positive side. The only point remaining common in all intervals as \(n\) approaches infinity is \(0\). Thus the intersection is \{0\}, which is not open because it includes its boundary without any surrounding interval.
03

Understanding Closed Sets and Unions

A closed set on the real line includes its boundary points. The intervals \[\frac{1}{n}, 1 - \frac{1}{n}\] are closed because they include the boundary points \(\frac{1}{n}\) and \(1 - \frac{1}{n}\). When we calculate their infinite union, we need to assess if the resulting set contains only boundary points or a range.
04

Solving the Union Example

Starting from \(n=2\), the intervals \[\frac{1}{n}, 1 - \frac{1}{n}\] expand as n increases. Eventually, this union covers all points in the open interval \(0, 1\). It does not include the boundary points \(0\) and \(1\) because there is no \(n\) that would include these exact endpoints in any interval. Hence, the set \(0, 1\) is not closed.

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Key Concepts

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

Open Sets
In topology, an open set is one that does not contain its boundary points. A familiar example on the real line is an open interval, such as (-1, 1). Open intervals don't include their endpoints, making them distinct from closed intervals. They can either be expressed in the form (a, b) where 'a' and 'b' are real numbers, or expanded to infinite cases like (-∞, b) or (a, ∞).

Open sets have an interesting interaction when intersected. If you take an infinite intersection of open sets, the result might not itself be open. For instance, consider the open interval (-1/n, 1/n) for each natural number n. As n grows, this interval squeezes towards the center. If you continue infinitely, the only remaining point common to all these intervals is the single point {0}. This resulting set {0} is not open because if it were, there would need to be some interval around it that lies entirely within an open set, which isn't possible as it contains a boundary point without an internal interval rich in points.
Closed Sets
Closed sets, in contrast to open sets, include all boundary points in the given interval. Consider the interval [1/n, 1 - 1/n] for each natural number n. Each of these intervals includes its endpoints, making them closed. If these endpoints are part of the set, then by definition of closed sets, the set would harness its boundary.

When thinking about closed sets, an interesting case arises when you explore their infinite unions. Normally, closed sets are well-defined by the inclusion of their boundary points, but the union of infinitely many such sets does not guarantee closure. For example, taking the union of [1/n, 1 - 1/n] for n starting from 2 up to infinity, the union broadens enough to eventually capture every point in the open interval (0, 1). However, it does not include 0 or 1 because there is no single interval that has these endpoints. Thus, the infinite union results in an open interval (0, 1), showing that the infinite union of closed sets can sometimes be open.
Infinite Intersections
Infinite intersections in topology are quite intriguing. When dealing with open sets, an infinite intersection reduces the originally open nature of sets quite drastically. Consider this: you take an infinite intersection of open intervals of the form (-1/n, 1/n), hence evaluating how they behave as n increases.

As n becomes exceedingly large, the intervals fast approach the single point 0. Thus, i.e., \(igcap_{n=1}^{ ext{∞}}(-1/n, 1/n) = \{0\}\). This is not open in the topological sense, since an open set must not include its boundary points, and 0 here lacks surrounding points to form an internal interval. Therefore, it's pivotal to note that the infinite intersection of open sets can collapse down to a non-open set, often just a point in such cases.
Infinite Unions
In theoretical mathematics, infinite unions can lead to quite unexpected outcomes, particularly when working with closed sets. To elucidate this, let's examine the closed intervals [1/n, 1-1/n].

If we continue to take the union of many such intervals starting from n=2, we start packing more points within each subsequent interval. This sequential union spans further and further towards the boundaries of (0, 1). Eventually, the union captures nearly everything in (0, 1).

Significantly, this expansive union of closed intervals [1/n, 1-1/n] between n = 2 to ∞ do not converge to a closed set but rather an open interval \((0, 1)\). This example reveals how the infinite union of closed sets can deviate from closure by being unable to secure boundary points, showing how flexibility in one scenario does not always sustain in another when extending to "infinite" scenarios.

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