# Topology on Eculidean n-space(ℝ^n)

by davechrist36
Tags: eculidean, nspaceℝn, topology
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,566 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.
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,566 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.