# Topology on Eculidean n-space(ℝ^n)

1. Jan 31, 2013

### davechrist36

Hey guys in fact here is my first time to have interaction over this forum!

I've already read how one can show the topology in ℝ(real Line) which is usual called standard topology fulfill the three condition fro to be topology. however,

I want to make inquiry on how can i proof whether the topology in ℝ^n(Euclidean n-space) satisfying the condition of topology, thus making topology ?

Last edited: Jan 31, 2013
2. Jan 31, 2013

### Number Nine

...what? You want to prove that "the topology" on Rn is a topology?

3. Jan 31, 2013

### HallsofIvy

Staff Emeritus
I think that what davechrist37 is saying is that he wants to prove that the collection of what are called "open sets" as normally defined on Rn is a topology. Of course, those are defined by the metric $d(x,y)= \sqrt{(x_1-y_1)^2+ (x_2- y_2)^2+ \cdot\cdot\cdot+ (x_n- y_n)^2}$ so it is only really necessary to prove that that is a metric. I believe that is typically done in any introductory Analysis course.

4. Feb 1, 2013

### davechrist36

thanx guys for you immediate concern. in fact i was looking for how the collection of open sets over Euclidean-n space forms topology; however right yesterday I got a material w/c help me how can I go thru the proof of that. let me glance ovr that and i'll reflect it here again.

5. Feb 1, 2013

### davechrist36

Re: Homeomorpism on Topology

let B^(n )∁ R^n denote the open unit ball in R^n with center at the orgine.i.e
B^(n )={x∈R^n:|x|<1}.Then how I can show or Prove the map f:R^n-→B^(n ) given by
f:x⟼x/(1+|x| ) ϵB^(n )
is well defined and gives a homeomorphism B^(n )≅ R^n

6. Feb 1, 2013

### HallsofIvy

Staff Emeritus
That last is a completely different question, and much harder, from your first question!
That, at least is relatively straight forward. Given a set X, a topology for X is a collection of subsets of X, T, such that:
1) X is in the collection.
2) The empty set is in the collection.
3) The union of any sets in the collection is also in the collection.
4) The intersection of any finite number of sets in the collection is also in the collection.

The "standard topology" for Rn is the collection of open sets where a set, A, is open if and only if "for ever x in A, there exist a number $\delta> 0$ such that the ball, $B_\delta(x)$, defined as $\{y | d(x,y)< \delta$ is a subset of A". Here d(x, y) is the "standard metric" on Rn: if $x= (x_1, x_2, ..., x_n)$, and $y= (y_1, y_2, ..., y_n)$, then $d(x,y)= \sqrt{(x_1- y_2)^2+ (x_2- y_2)^2+ \cdot\cdot\cdot+ (x_n- y_n)^2}$.

Now, suppose x is in $\cup \{U_i\}$ where $\{U_i\}$ is a collection of open sets A. Then there exist some specific $U_i$ containing x. Since $U_i$ is open, there exist $\delta> 0$ such that $B_\delta(x)\subset U_i$. But if every point of $B_\delta(x)$ is in $U_i$j it is certainly in the union so $B_\delta(x)\subset \cup \{U_i\}$ and so $\cup \{U_i\}$ is an open set.

Intersection is a little trickier and why we need "finite". Suppose $x\in \cap U_i$. Then $x\in U_i$ for all i. Since every $U_i$ is open, there exist $\delta_i> 0$ such that $B_{\delta_i}(x)\subset U_i$ for every i. Here's where we need "finite". Since the set of all such $\delta_i$ is finite, there exist a smallest $\delta_k$. Then $B_{\delta_k}(x)$ is a subset of all such $B_i(x)$ and so is in every $U_i$ and so in their intersection. $B_k(x)$ for that k is in $\cap \{U_i\}$. Since x could be any point in $\cap\{U_i\}$, $\cap \{U_i\}$ is open.

Now that we have shown that the union of any sub-collection is also in the collection, to show that the entire space, A, is in the collection, take the union of all sets in the collection. To show that the empty set is in the collection, take the union of of the empty subcollection.