Question

Let \(G_{1}\) be an open, dense set in a normed linear space \(X\). Prove that for any \(x \in X\) and \(r>0\), the set \(B_{r}(x) \cap G_{1}\) contains a ball. Deduce that if \(G_{1}\) and \(G_{2}\) are open and dense in \(X\), then \(G_{1} \cap G_{2}\) is also open and dense in \(X\).

Short answer

Question: Prove that for any x in a normed linear space X and r>0, if G1 is an open, dense set in X, the intersection between the open ball (center x with radius r) and G1 contains another ball. Deduce that if G1 and G2 are both open and dense in X, the intersection of G1 and G2 is also open and dense in X. Answer: To prove that the intersection B_r(x) ∩ G1 contains a ball for any x in X and r>0, we showed that (1) if x is a point in G1, there exists a ball contained in both B_r(x) and G1, and (2) if x is a limit point of G1, then for any t>0, B_t(x) intersects G1, and B_r(x)∩G1 contains a ball. Based on this, we deduced that if G1 and G2 are open and dense sets in X, their intersection G1 ∩ G2 is also open and dense in X by showing that G1 ∩ G2 is (1) open and (2) dense in X.

Step-by-step solution

Recall definitions

Let's recall some important definitions for this exercise: 1. Open set: A set G in a normed linear space X is open if for every x in G, there exists r>0 such that B_r(x) is a subset of G, where B_r(x) denotes an open ball centered at x with radius r. 2. Dense set: A set G in a normed linear space X is dense if the closure of G is X. In other words, every point of X is either a point of G or a limit point of G. 3. Ball in normed linear space: Given a normed linear space X, a ball centered at x with radius r is the set B_r(x)={y in X: ||x-y||<r}, where ||.|| is the norm in X.

Prove the intersection set contains a ball

Let x in X and r>0. We need to prove that B_r(x)∩G1 contains a ball. Since G1 is dense in X, the closure of G1 is X. Therefore, x is either a point of G1 or a limit point of G1. - If x is a point of G1, then G1 is an open set so there exists s>0 such that B_s(x) is a subset of G1. Let s_min = min{r, s}. Then B_s_min(x) is a ball contained in both B_r(x) and G1, so B_r(x)∩G1 contains a ball. - If x is a limit point of G1, then for any t>0, B_t(x) intersects G1. Particularly, B_r(x) intersects G1. Let y be a point in B_r(x)∩G1. Since G1 is open, there exists u>0 such that B_u(y) is a subset of G1. Let u_min=min{u,r-||x-y||}. Then, B_{u_min}(y) is a ball contained in both B_r(x) and G1, so B_r(x)∩G1 contains a ball.

Deduce G1 ∩ G2 is open and dense

Let G1 and G2 be open and dense sets in X. We will show that their intersection G1 ∩ G2 is also open and dense in X. - To show G1 ∩ G2 is open: Let z be in G1 ∩ G2. Since G1 and G2 are open, there exist r1, r2 > 0 such that B_{r1}(z) is a subset of G1 and B_{r2}(z) is a subset of G2. Let r = min{r1, r2}. Then B_r(z) is a subset of both G1 and G2, so B_r(z) is a subset of G1 ∩ G2. Therefore, G1 ∩ G2 is open. - To show G1 ∩ G2 is dense: Let x be in X. Since G1 and G2 are dense in X, their closures equal X. Then, x must be in the closure of G1 ∩ G2. To see why, given any r>0, B_r(x)∩G1 contains a ball, by Step 2, so B_r(x)∩G1 is nonempty. Furthermore, B_r(x)∩G2 is nonempty since G2 is dense in X. Since B_r(x)∩G1 is nonempty and B_r(x)∩G2 is nonempty, it follows that B_r(x)∩(G1∩G2) is nonempty, i.e., x is in the closure of G1 ∩ G2. As this holds for any x in X, G1 ∩ G2 is dense in X. By proving both, we have deduced that if G1 and G2 are open and dense in X, then their intersection G1 ∩ G2 is also open and dense in X.

Key concepts

Open Sets

Understanding open sets is fundamental in mathematical analysis and topology. An open set within a normed space, such as a normed linear space, is a collection of points characterized by a particular openness around each point in the set. For every point in an open set, you can find an open ball that fits entirely within the set and contains our initial point at its center.

We can define an open ball, denoted as B_r(x), centered at a point x with a radius r as the set of all points that have a distance less than r from x. This concept of an open ball speaks intuitively to the idea of 'room to wiggle'. If you pick any point in an open set, you can wiggle around that point without leaving the set. By providing a vivid picture, students might find it easier to imagine this space around the point, which is free from the set's boundary.

Ball in Normed Space

A normed space comes equipped with a function called 'norm', which measures the length or size of a given vector within that space. The ball in a normed space is an essential concept that can be thought of as a generalization of the idea of a circle (in two dimensions) or a sphere (in three dimensions) to any number of dimensions.

When representing this geometrically, envision blowing up a balloon from a point x in our space, with the balloon's size dictated by a predefined radius r. The inside surface of this balloon represents the boundary of the ball, B_r(x). It consists of all points that are within a given distance, the radius r, from the center x, without including the surface itself, which makes it an 'open' ball.

Intersection of Dense Sets

The notion of dense sets intersects with the concept of closure in topology. A set is dense in a space if its closure is the whole space, which means that within any region of the space, no matter how small, you will find points from the dense set. Visualize a dense set like a fine mist sprayed in a container: although the individual droplets are tiny, they spread out to fill every bit of space.

When considering the intersection of dense sets, it's like overlapping two such mists. The intersection, despite involving two different sets, remains dense within the original space. This might seem a bit paradoxical at first, because the intersection could be thought to be thinner than the individual sets; however, because both sets are so thoroughly spread throughout the space, their overlap continues to reach everywhere in that space. Knowing this helps students understand that even when we 'narrow down' our focus by intersecting dense sets, they still reach out to touch every part of our space.