How can I prove that the set A in metric space (X, D) is closed if A complement is open?
Roughy speaking, another definition of closed sets (more common in analysis) is that A contains the limit point for every convergent sequence of points in A. Every point in X must be in A or A’s complement, but not both. Consider the limit point, p(lim) for any convergent sequence of elements in A (for every epsilon >0, there exists an N such that D(p(lim)-p(n)) < epsilon for all p(n) in A when n>N). A is closed if p(lim) is also in A. Therefore p(lim) cannot be in A’s complement. And, for all sequences of p in A’s complement that approach p(lim), the limiting element cannot be in A’s compliment. Therefore, A’s complement is an open set. A contains its boundary elements, and A’s complement cannot contain its boundary elements, so it is open.Example: X = (0, 1). A=[0.467, 1). A is closed (within X), because for all sequences of points in A that approach 0.467, the limiting value of 0.467 is in A. A’s complement is open because for all sequences in A’s complement that approach 0.467 (from below), the limiting value of 0.467 is not in A’s complement. The boundary in this case is simply one point, 0.467, and that boundary point is in A, not A’s complement. (The point 1 is not in X, so we say A is closed even though it does not contain 1. A is closed relative to X.)You will need to make all this more formal for full credit.Edit: I hate autocorrect.
If (M , d) is a metric space so that every subset of M is closed, how do you prove that a subset of M is compact, if and only if it's finite?
If every subset is closed, then every singleton subset is the complement of a closed set, and therefore open.If there is only a finite number of singleton subsets, there isn't really anything to prove; every open cover is already finite. Otherwise, pick some infinite set of points, and cover it with the corresponding singleton open sets. Bingo! The cover is infinite, but deleting any of those open sets means that the corresponding single point is no longer covered. There is no finite sub-cover, and your set is not compact. This applies to any set your start with, including the entire space.Notice that I haven't used the metric properties at all; the argument holds for any point-set topology. If you want a metrified version, just consider that in order to be able to isolate a single point there must be an open ball around it which includes no other points; and then your have a metric-style open set containing just the single point.
An open ball is an open set. But is the converse true, that is, is an open set an open ball? Provide Ex.?
Is an open set an open ball? If not, provide an example. Also a closed ball is a closed set. But, is a closed set a closed ball? If not, provide example. This is for Real Analysis, not Topology. Thanks,
What is space-time interval?
As was said above, the space-time interval between two events (3 space coordinates, 1 time coordinate) is the analogue to the distance between two points in R^3 when one adds a time axis. In flat spacetime, the interval is the dI that satisfies (dI)^2 = (dx)^2 + (dy)^2 + (dz)^2 - (cdt)^2. Invariants of this form are called metrics. You can derive this metric from the postulates of special relativity, specifically that the speed of light in vacuum is an invariant. Curved manifolds will have different metrics; just like the distance between two points on a sphere (as measured on the sphere) does not follow the Pythagorean theorem. The Schwarzschild metric is a metric for the spacetime around a massive body. See http://en.wikipedia.org/wiki/Schwarzschi... .
Is there an intuitive definition of the open ball?
Is there an intuitive definition of the open ball?All the points strictly less than a certain distance from a given point.Formally in math-speak, given a set, [math]S[/math], with a metric (or distance), [math]d[/math], the open balls about a point [math]x\in S[/math], are the sets:[math]\quad\{y\in S\mid d(x,y) < c\}[/math]for various constants [math]c\in\mathbb R[/math].The balls are called “open” because given any point in an open ball you can find a (smaller) ball around the point that is wholly contained in the original ball.The open balls of any metric space provide a natural topology for the space.Physical space with our usual straight-line definition of distance is a metric space in which a tennis ball (without its boundary) is indeed an open ball – with its boundary (probably necessary for a game of tennis) it would be a closed ball.