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Expert-verifiedFound in: Page 930

Book edition
1st

Author(s)
Peter Kohn, Laura Taalman

Pages
1155 pages

ISBN
9781429241861

Problem Zero: Read the section and make your own summary of the material.

Summary of the section 12.2 includes Open sets and closed sets, Limit of a function and Continuity of a function.

Section 12.2 covers the concepts of open sets, closed sets, the limit of a function, and the continuity of a function.

To understand the concept of open sets, we first need to understand what an open disk or ball is. An open disk in $$\mathbb{R}^ 2 $$ is a subset of the form $$\{ (x,y) | (x-x_{0})^2+(y-y_{0})^2 < \epsilon \}$$A open ball in $$\mathbb{R}^ 3 $$ is a subset of the form $$\{ (x,y,z) | (x-x_{0})^2+(y-y_{0})^2 +(z-z_{0})^2 < \epsilon \}$$A subset $$S$$ of $$\mathbb{R}^ 2 $$ or $$\mathbb{R}^ 3 $$ is said to be open if, for every point $$(x,y)$$ or $$(x,y,z)$$ in $$S$$there exists a open disk $$D$$ or ball $$B$$ such that $$D \in S$$ or $$B \in S$$A closed set is a set whose complement is open. i.e for $$A$$ to be a closed set, $$A^{c}$$ must be open.

Let $$S$$ be a subset of $$\mathbb{R}^ 2 $$ or $$\mathbb{R}^ 3 $$ A point $$(x,y)$$ or $$(x,y,z)$$ in $$S$$ is said to be a boundary point if every open disk or ball containing it intersects both $$S$$ and $$S^{c}$$The set of all boundary points of $$S$$ is called the boundary of $$S$$A subset of $$\mathbb{R}^ 2 $$ or $$\mathbb{R}^ 3 $$ is said to be bounded if it is a subset of some open disk or ball in $$\mathbb{R}^ 2 $$ or $$\mathbb{R}^ 3$$ A subset which is not bounded is said to be unbounded.

Let $$f$$ be a function of two or more variables. The limit of $$f$$ at $$a$$ is $$L$$ if, for every $$ \epsilon > 0$$there exists $$ \delta >0$$such that $$ | f-L|<\epsilon$$ whenever $$0 < |x-a|<\delta$$Then we can write \[ \lim_{x\to a} f(x) = L \]In the case where $$f$$ is a function is two variables, $$ x= \langle (x,y) \rangle $$ and $$ a= \langle (a,b) \rangle $$Therefore \[ \lim_{x\to a} f(x) = L \] means $$ | f(x,y)-L|<\epsilon$$ whenever $$0 < \sqrt{(x-a)^2+(y-b)^2}<\delta$$. In the case where $$f$$ is a function is three variables, $$ x= \langle (x,y,z) \rangle $$ and $$ a= \langle (a,b,c) \rangle $$. Therefore \[ \lim_{x\to a} f(x) = L \] means $$ | f(x,y,z)-L|<\epsilon$$ whenever $$0 < \sqrt{(x-a)^2+(y-b)^2+(z-c)^2}<\delta$$

Algebra of limits of function in one variable can be applied directly to the functions of two or more variables.

The limit of a function in two or more variables exists if and only if the the limit is same for every path containing the given point. i.e. the limit \[ \lim_{x\to a} f(x) = L \] exists if and only if \[ \lim_{x\to a} f(x) = L \] for every path $$C$$ containing $$(a,b)$$ in open set $$S$$.

A function $$f$$ in two or more variables defined over an open set $$S$$ is said to be continuous at a point $$a$$ in $$S$$ if \[ \lim_{x\to a} f(x) = f(a) \]. Also, $$f$$ is continuous on $$S$$ if it is continuous on every point in $$S$$ A function is said to be continuous everywhere if it is continuous at every point of it's domain.

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