Question on the properties of a manifold

[majutsu]

According to my text, a manifold should be 1) Hausdorff (that is t-2 separable, so there are disjoint open sets which are neighborhoods for any two points x and y), 2) locally euclidian (that there is a neighborhood U of a point x that is homeomorphic to an open subset U' of Rn (the RxR...xR cartesian product) and 3) has a countable basis of open sets.

In most books, when the set out to illustrate something is a manifold, they usually explicitly show the locally euclidian character. Then the embedding of that manifold in En (euclidian n-space) is used to assume the hausdorff and countable basis (paracompact) requirements. The hausdorff assumption, I have no trouble with, as Rn clearly meets that condition as I understand it above. But I don't assume that En or Rn have a countable basis of open sets.

Can someone prove to me that Rn has a countable basis of open sets?

 

[mathwonk]

try to prove it yourself. start with R. take the most natural countable set of center points i.e. the rationals., and take the most natural countable collection of open balls, i.e. rational radius intervals.

a piece of advice however:

these technical properties have no importance whatever, except that they are true; i.e. if you cannot prove them, but are willing to assume them, you are still going to get everything possible from the subject that is of importance.

of course i understand a curious poerson has difficulty by passing a puzzling statement.

 

[M98Ranger]

Yes, it can be proven that Rn has a countable basis of open sets. This is one of the properties of a topological space that is often assumed and not explicitly stated.

To prove this, we first define a basis for a topological space. A basis for a topological space X is a collection B of open sets such that every open set in X can be written as a union of elements of B. In other words, the elements of B "generate" all the open sets in X.

Now, let's consider the standard topology on Rn, which is the topology induced by the standard metric on Rn. In this topology, the open sets are defined as follows: a set U is open if and only if for every x in U, there exists a positive real number r such that the open ball of radius r centered at x is completely contained in U.

Using this definition, we can construct a basis for Rn as follows: for every positive integer n, we can consider the set of all open balls in Rn with rational radii and rational center coordinates. This set, let's call it Bn, is countable since the rationals are countable.

Now, we claim that B = {Bn | n is a positive integer} is a basis for Rn. To see this, let U be any open set in Rn. Then for every x in U, there exists a positive real number r such that the open ball of radius r centered at x is completely contained in U. Since the rationals are dense in the reals, we can always find a rational number r' that is arbitrarily close to r. Therefore, the open ball of radius r' centered at x is also contained in U. This means that for every x in U, we can find an element of B that contains x and is completely contained in U. Hence, U is a union of elements of B and B is indeed a basis for Rn.

Therefore, Rn has a countable basis of open sets, as claimed.