Question

Regard \(Q\), the set of all rational numbers, as a metric space, with \(d(p, q)=|p-q|\). Let \(E\) be the set of all \(p \in Q\) such that \(2

Short answer

Yes, the set \(E\) is closed and bounded in \(Q\). 2. Is the set \(E\) compact? No, the set \(E\) is not compact. 3. Is the set \(E\) open in \(Q\)? No, the set \(E\) is not open in \(Q\).

Step-by-step solution

Define closed and bounded sets in a metric space

A set is closed in a metric space if it contains its limit points. We can express this definition using sequences and convergence. Let us first prove that \(E\) is bounded in \(Q\). Recall that a set is bounded if there exists an upper and lower bound for it. For \(E\), let us take lower bound \(a = \sqrt{2}\) and upper bound \(b = \sqrt{3}\). Since \(a^2 = 2\) and \(b^2 = 3\), by definition, all elements \(p \in E\) satisfy the condition \(2<p^2<3\). For any \(p\in E\), we can find \(q\in Q\) such that \(a<q<p\). Similarly, for \(p^2<3\), we can find \(r\in Q\) such that \(p<r<b\). Hence, we have a lower and upper bound for the set, making it bounded. Now we need to check the closure of the set \(E\). Since \(E\) is a subset of \(Q\), we will be checking if it is closed in its parent set \(Q\). We can do this by verifying if the limit points of sequences are present in \(E\).

Prove that \(E\) is closed

Let's consider a sequence of rational numbers \((p_n)\) in \(E\) that converges to a limit, say \(L\). We know that \(|p_n - L| \to 0\) as \(n \to \infty\). We need to show that \(L \in E\). Since each \(p_n\) is an element of \(E\), we have \(2 < p_n^2 < 3\) for all \(n\). Taking the limit as \(n \to \infty\), we have: $$2 < \lim_{n\to\infty} p_n^2 < 3$$ which implies that \(2 < L^2 < 3\). Hence, \(L \in E\), which means that \(E\) is closed.

Define compactness and show that \(E\) is not compact

A set is compact if every open cover has a finite subcover. In simple words, we can cover the entire set using a finite number of smaller open sets. In the context of \(E\), let's consider the sequence of intervals given by: $$I_n = \left(p-\frac{1}{n}, p+\frac{1}{n} \right)$$ where \(n = 1, 2, 3,\dots\) and \(p\in E\). Since every \(p \in E\) lies in one such interval \(I_n\), these intervals form an open cover for the set \(E\). However, due to the density of rational numbers, we cannot find a finite subcover of these intervals that would still cover the entire set \(E\). Therefore, \(E\) is NOT compact.

Investigate the openness of \(E\) in \(Q\)

A set is open if every point in the set has a neighborhood completely contained in the set. By definition, an open neighborhood around a point \(p\) in a metric space is a set that contains points \(q\) such that the distance between \(p\) and \(q\) is less than a specified value. Let \(p \in E\) be any element, and consider any neighborhood of \(p\) with radius \(\epsilon > 0\). The openness of \(E\) implies that there exists a \(q \in E\) such that \(p - \epsilon < q < p + \epsilon\). Consider \(q' = \frac{p+q}{2}\). This value lies in the neighborhood of \(p\), and we have \(p - \epsilon < q' < p + \epsilon\), but since \(q'\in Q\), we can't guarantee that \(q'\) satisfies \(2<q'^2<3\). That means \(q'\) can be outside the set \(E\), implying that the neighborhood isn't completely contained in \(E\). Thus, \(E\) is NOT open in \(Q\). In conclusion, the set \(E\) is closed and bounded in \(Q\), but it is not compact, and it is not open in \(Q\).

Key concepts

Bounded Sets in Metric Spaces

Understanding boundedness in metric spaces is crucial for analyzing the structure of these mathematical entities. In the context of the set \(E\), boundedness refers to the idea that the set has finite 'limits.' This essentially means that we can draw a 'mathematical circle,' or more accurately, specify a range within which every element of the set must lie.

The rational number set \(E\) from our exercise, containing numbers whose squares are strictly between \(2\) and \(3\), is a perfect example. By demonstrating that \(\sqrt{2} < p < \sqrt{3}\) for any \(p\) in \(E\), we establish that there is a limit to how small or large the elements of \(E\) can be. As a student, visualizing these boundaries is a stepping stone to understanding more complex concepts in analysis.

Closed Sets in Metric Spaces

When we talk about closed sets, think of a 'mathematically sealed border' that doesn't let any of its limit points escape. For any sequence of numbers inside \(E\) that approaches a certain number as you go infinitely further in the sequence, that number will also be in \(E\).

This concept might seem tricky, but it's like saying that if you are walking towards your home and every step takes you closer, eventually you'll reach home — home being the closed set. In our example with the set \(E\), take any sequence of rational numbers that inch closer and closer to a limit. Since the square of the limit still falls between \(2\) and \(3\), the limit is also part of \(E\), making it a closed set. This provides a solid ground for students to understand more intricate ideas such as continuity and convergence.

Compactness in Metric Spaces

To grasp the concept of compactness, imagine trying to cover a shape with a finite number of patches so that the entire shape is covered without any gaps or overlaps. If you can do this, the shape (or set) is compact. It's a bit like putting finite pieces of a puzzle together to complete a picture.

The set \(E\) throws a curveball because while it looks like it can be covered with small open intervals, rational numbers are dense, and you'd actually need an infinite number of patches (or intervals). Hence, no matter how hard you try, \(E\) will stubbornly refuse to be compact. This is a nuanced topic that illustrates the interplay of finiteness and covering in metric spaces, key to understanding various theorems in real analysis and topology.

Open Sets in Metric Spaces

An open set is like a party that's only for an exclusive group. If you're part of the set, everyone around you is also part of the set — there are no outsiders. However, the set \(E\) in our exercise doesn't meet this criterion because it excludes certain points that are incredibly close (neighbors, if you will) to the numbers in \(E\). These 'almost included' numbers have rational squares less than \(2\) or greater than \(3\), hence they're not invited to our exclusive \(E\) party.

This exclusion is the key to understanding why \(E\) is not open. If you take any point in \(E\), and look closely around it, you'll find points that don't belong to \(E\), making the neighborhood not fully contained in the set. Recognizing the openness of a set is critical for studying continuous functions and the structure of spaces in higher mathematics.