Show that every metric space can be made a topological space.?
Call the metric space M with the metric d. For any r > 0 and x ∈ M, let B(x,r) denote the open ball of radius r centered at x. In other words, B(x,r) = {y ∈ M | d(x,y) < r}. Let ℬ denote the collection of all open balls: ℬ = {B(x,r) | x ∈ M, r > 0}. I claim that ℬ is a base for a topology on M. To show that ℬ is a base, we need to verify two things: (1) ℬ covers M (2) if B(x,r) and B(y,s) in ℬ, and if z ∈ B(x,r)∩B(y,s), then there is a ball B(z,t) contained in B(x,r)∩B(y,s). proof: (1) For any point x ∈ M, we know that x ∈ B(x,1), and B(x,1) ∈ ℬ. Hence, the sets in ℬ cover all of M. (2) Suppose z ∈ B(x,r)∩B(y,s). Let t = min{r - d(x,z), s - d(y,z)}. Since z ∈ B(x,r) and z ∈ B(y,s), then d(x,z) < r and d(y,z) < s, so the number t we just defined is positive. Now, we want to check that B(z,t) ⊆ B(x,r)∩B(y,s). For any point w ∈ B(z,t), we have d(z,w) < t, so the triangle inequality tells us: d(x,w) <= d(x,z) + d(z,w) < d(x,z) + t <= d(x,z) + (r - d(x,z)) = r, d(y,w) <= d(y,z) + d(z,w) < d(y,z) + t <= d(y,z) + (s - d(y,z)) = s. Hence, w ∈ B(x,r) and w ∈ B(y,s), which means B(z,t) ⊆ B(x,r)∩B(y,s). Thus, ℬ is a base, which means ℬ generates a topology T on M: T = {Ø} U {all unions of elements of ℬ}. And for any nonempty subset U of M, we have: U ∈ T (i.e. U is "open") ⇔ U is a union of open balls ⇔ for any x ∈ U, there is an open ball B such that x ∈ B ⊆ U ⇔ U is open in the usual sense as defined when discussing metric spaces So the topology T on M agrees with the notion of "open set" that is discussed with metric spaces.
Basis of Topological Space?
A basis for the usual topology on Euclidean space is the open balls. An open ball of radius r > 0 centered at a point x, is defined as the set of all y in R^n whose distance from x is strictly smaller than r. By the way the topology on R^n is defined, these open balls clearly form a basis. Given any metric space, there is a natural way of viewing it as a topological space. This natural way involves declaring the collection of open balls in the metric space as a basis. To see that this gives a well-defined topology, we need to check that the collection of open balls satisfies the conditions to be a basis for a topological space. This follows from the conditions for a metric space, including nonnegativity, symmetry and the triangle inequality. Regarding the metric space, define the metric as the boolean = 0 if the points are the same and 1 = if the points are different. (So, every point is exactly 1 unit away from every other point.) Now it should be clear that the union of all neighborhoods is the union of all points and the intersection of any collection of neighborhoods is an intersection of discrete points.
Can you help me with a Metric Spaces proof?
Suppose d1 and d2 are two metrics on S. Let C be a positive real number such that d1(x,y)≤Cd2(x,y) for each x and y in S. Let G be a subset of S. Prove that if G is an open set in (S,d1), then G is also an open set in (S, d2). I hate these really "obvious" proofs. The statement makes sense to me but I have no idea how to show it. Any help is appreciated, even if you can just tell me how to start off!