Prove Rn is a Hausdorff space?

Question

How do you prove that Rn is a Hausdorff space?
Rn = Real with dimension n

Answer 1

We have to prove that, if x and y are distinct elements of R^n, then there are disjoint neighborhoods Bx of x and By of y. Since in R^n open balls are neighborhoods, it suffices to show the assertion supposing Bx and By are open balls centered at x and y, respectively.

Let r = ||x - y||.(|| || is the Euclidean norm) Then r>0, for x <>y. Let Bx be the open ball of radius r/2 centered at x and By the open ball of radius r/2 centered at y. If u is in Bx, then ||u - x || < r/2. Then, according to the triangle inequality (one of its variants), we have ||y - u|| = || (y- x) - (u - x) || >= | ||y - x|| - ||u-x|| | = | r -r/2| = r/2, so that ||y - u|| > r/2. This shows u is not in By, that is, no element of Bx is in By, the same as to say Bx and By are disjoint.

Since this holds for every pair of distinct elements of R^n, it follos R^n is a Hausdorff space.

Actually, every metric space is a Hausdorffr space. The proof is exactly the same, all you have to do is replace the Euclidean norm by the distance function defined in the metric space.

Answer 2

any metric space is a hausdorff space. If you have two points a certain distance apart, you can calculate the radius for a ball around each so that the balls are disjoint.

Answer 3

"Gf is a subset of XxY and XxY is a Hausdorff area so Gf is a Hausdorff area. " Y is Hausdorff area, XxY isn't inevitably Hausdorff area. as a result Gf isn't inevitably Hausdorff area "enable (x1,y1) be a decrease component to Gf. would desire to coach that (x1,y1) is in Gf. " perfect "because of the fact (x1,y1) is a decrease component to Gf, there exists an open set U in Gf such that (x1,y1) is in U and a few element (x2,y2) is in U additionally." If, as you're saying, U is in Gf and (x1,y1) is in U, then (x1,y1) could be in Gf, isn't it? so as that could practice the subject. the perfect definition of decrease element is that " Any open set U in XxY such that (x1,y1) is in U will contain additionally a element in Gf." "Gf is a Hausdorff area so for all (x2,y2),(x3,y3) in Gf there exists open instruments U,V in Gf such that (x2,y2) is in U and (x3,y3) is in V and the intersection of U and V is the null set" perfect "(x1,y1) is in Gf on condition that's in some open set U in Gf and the intersection of this set with another arbitrary open set in Gf is the null set. " That is senseless usual, you probably did no longer use at each and all the definition of Gf. would not this make the info suspect? No offence.

Answer 4

You must show that any two distinct points in Rn are separated by neighborhoods.