In the preceding posts, we have seen some properties of real numbers.
(1) A cut of a real number line satisfies the condition of Dedekind cut in which there exists $\max S$ and does not $\min T$ , or there exists $\min T$ and does not $\max S$ .
(2) Bounded monotone real number sequences will converge.
(3) If a set of real numbers is bounded above or below, then it has a supremum or a infimum.
(4) If $a$ and $b$ are positive real numbers, then there is an $n\in \mathbb{N}$ such that $na>b$. (It is called the Archimedean property. ) In an addition, nested intervals have a limit real number.
By the strict proofs, we will get that these four properties are equivalent. Therefore, we are able to adopt which properties as an axiom of completeness.
When four properties are apposed, you will feel a similar work, won't you?
This is a last post of 2013. Best wishes for a happy new year 2014 ! !
2013/12/29
2013/12/23
one-sided continuity of a funtion
The definition of the continuity of a function is as follow.
The function $f$ is continuous, if for any $\epsilon>0$ there is a $\delta>0$ such that
\[ |f(x)-f(c)|<\epsilon \]
for every points $x$ for which $|x-c|<\delta$ .
It means that $\lim_{x\rightarrow -c}f(x)=f(c)$ and $\lim_{x\rightarrow +c}f(x)=f(c)$ .
By the preceding definition of one-sided limits, we are able to expand the continuity or discontinuity of a function.
A function $f(x)$ is continuous from the right at point $c$ if $\lim_{x\rightarrow +c}f(x)=f(c)$ .
Similarly, A function $f(x)$ is continuous from the left at point $c$ if $\lim_{x\rightarrow -c}f(x)=f(c)$ .
There are three kinds of discontinuity at point $c$ .
(1)Removable discontinuity : $\lim_{x\rightarrow c}f(x)$ exists. But $\lim_{x\rightarrow c}f(x)\ne f(c)$ . If we can redefine the function $f$ except a point $c$ , the discontinuity will be removed.
\[ (\mbox{example})\quad f(x)=\frac{x^2+x-2}{x-1} \]
(2)Jump or Step discontinuity : One-sided limits exists.
\[ (\mbox{example})\quad f(x)=\left\{ \begin{array}{cccc}
-x^2 & (x<0)& & \\
x^2+1 & (x\geq 0)& &
\end{array}
\right. \]
This example means $f(x)$ is continuous from the right at point 0, but not from the left.
(3)Infinite or essential discontinuity : One or both of the one-sided limits do not exist or infinite.
\[ (\mbox{example})\quad f(x)=\frac{1}{x-1} \]
The function $f$ is continuous, if for any $\epsilon>0$ there is a $\delta>0$ such that
\[ |f(x)-f(c)|<\epsilon \]
for every points $x$ for which $|x-c|<\delta$ .
It means that $\lim_{x\rightarrow -c}f(x)=f(c)$ and $\lim_{x\rightarrow +c}f(x)=f(c)$ .
By the preceding definition of one-sided limits, we are able to expand the continuity or discontinuity of a function.
A function $f(x)$ is continuous from the right at point $c$ if $\lim_{x\rightarrow +c}f(x)=f(c)$ .
Similarly, A function $f(x)$ is continuous from the left at point $c$ if $\lim_{x\rightarrow -c}f(x)=f(c)$ .
There are three kinds of discontinuity at point $c$ .
(1)Removable discontinuity : $\lim_{x\rightarrow c}f(x)$ exists. But $\lim_{x\rightarrow c}f(x)\ne f(c)$ . If we can redefine the function $f$ except a point $c$ , the discontinuity will be removed.
\[ (\mbox{example})\quad f(x)=\frac{x^2+x-2}{x-1} \]
(2)Jump or Step discontinuity : One-sided limits exists.
\[ (\mbox{example})\quad f(x)=\left\{ \begin{array}{cccc}
-x^2 & (x<0)& & \\
x^2+1 & (x\geq 0)& &
\end{array}
\right. \]
This example means $f(x)$ is continuous from the right at point 0, but not from the left.
(3)Infinite or essential discontinuity : One or both of the one-sided limits do not exist or infinite.
\[ (\mbox{example})\quad f(x)=\frac{1}{x-1} \]
2013/12/16
limits of a function (one-sided limits)
If a function $f(x)$ approaches $c$ as its argument $x$ approaches a point $x_0$ , the function $f$ is said to approach the limit $c$ . Now we shall address the following function.
\[ f(x)=\left\{ \begin{array}{cccc}
-x^2 & (x<0)& & \\
1& (x=0)& & \\
x^2 & (x>0)& &
\end{array}
\right. \]
As this function is not continuous, it is not easy for us to understand the limit of the function.
If $x$ approaches $0$ from the left-side hand, $f(x)\rightarrow 0$ . This result is also same in the case of $x$ approaching from the right-side hand. However, as $f(0)=1$ by its own definition, $\lim_{x\rightarrow 0} f(x)\ne f(0)$ .
We shall define two kinds of limit of a function. The limit of $f(x)$ in the case which $x\rightarrow x_0$ and $x<x_0 $ is said to the left hand limit and is written by
\[ \lim_{x\rightarrow -x_0}f(x) \]
If the right hand limit, then
\[ \lim_{x\rightarrow +x_0}f(x) \]
These are called one-sided limits. "left hand" is equivalent to "below" and "right hand" is equivalent to "above".
If for any $\epsilon>0$ there is a $\delta>0$ such that $|f(x)-c|<\epsilon$ whenever $0<x_0-x <\delta$ , the limit of $f(x)$ is $c$ as $x$ approaches $x_0$ from below. If the case is from above or right hand, the condition becomes $0<x-x_0<\delta$ .
In the above example, both the right hand and left hand limit of $f(x)$ at the point $0$ are not $f(0)$ unfortunately.
\[ f(x)=\left\{ \begin{array}{cccc}
-x^2 & (x<0)& & \\
1& (x=0)& & \\
x^2 & (x>0)& &
\end{array}
\right. \]
As this function is not continuous, it is not easy for us to understand the limit of the function.
If $x$ approaches $0$ from the left-side hand, $f(x)\rightarrow 0$ . This result is also same in the case of $x$ approaching from the right-side hand. However, as $f(0)=1$ by its own definition, $\lim_{x\rightarrow 0} f(x)\ne f(0)$ .
We shall define two kinds of limit of a function. The limit of $f(x)$ in the case which $x\rightarrow x_0$ and $x<x_0 $ is said to the left hand limit and is written by
\[ \lim_{x\rightarrow -x_0}f(x) \]
If the right hand limit, then
\[ \lim_{x\rightarrow +x_0}f(x) \]
These are called one-sided limits. "left hand" is equivalent to "below" and "right hand" is equivalent to "above".
If for any $\epsilon>0$ there is a $\delta>0$ such that $|f(x)-c|<\epsilon$ whenever $0<x_0-x <\delta$ , the limit of $f(x)$ is $c$ as $x$ approaches $x_0$ from below. If the case is from above or right hand, the condition becomes $0<x-x_0<\delta$ .
In the above example, both the right hand and left hand limit of $f(x)$ at the point $0$ are not $f(0)$ unfortunately.
2013/12/05
real number system 10 (Cauchy sequence)
In nested intervals, let $a_n$ be the increasing sequence from the left-hand side and $b_n$ be the decreasing sequence from the right-hand side. In the definition of the convergence the crucial condition was $|a_n-b_n|\rightarrow 0$ as $n\rightarrow \infty$ .
Given a sequence $c_n$ and consider a sequence $n_i$ of positive integers such that $n_1<n_2<\cdots $ . Then the sequence $c_{n_i}$ is called a sub sequence of $c_n$ .
It is clear that $c_n$ converges to $c$ if and only if every sub sequence of $c_n$ converges to $c$ .
A sequence $c_n$ is said to be a Cauchy sequence if for any $\epsilon>0$ there is a $N$ such that $|c_n-c_m|<\epsilon$ if $n,m\geq N$ . (It means $|c_n-c_m|\rightarrow 0$ but $[c_n,c_m]$ are not always nested. )
We need to know the theorem in which every convergent sequence is a Cauchy sequence. However in the theorem the limit is not explicitly involved.
Given a sequence $c_n$ and consider a sequence $n_i$ of positive integers such that $n_1<n_2<\cdots $ . Then the sequence $c_{n_i}$ is called a sub sequence of $c_n$ .
It is clear that $c_n$ converges to $c$ if and only if every sub sequence of $c_n$ converges to $c$ .
A sequence $c_n$ is said to be a Cauchy sequence if for any $\epsilon>0$ there is a $N$ such that $|c_n-c_m|<\epsilon$ if $n,m\geq N$ . (It means $|c_n-c_m|\rightarrow 0$ but $[c_n,c_m]$ are not always nested. )
We need to know the theorem in which every convergent sequence is a Cauchy sequence. However in the theorem the limit is not explicitly involved.
2013/12/01
real number system 9 (bounded monotone sequences)
In the method of nested intervals, we supposed the following proposition.
If a real number sequence increasing or decreasing monotonically is bounded, it will converge.
We shall prove it simply. A monotonical sequence is either
$(increasing)\quad a_1\leq a_2\leq\cdots\leq a_n\leq\cdots$ , or
$(decreasing)\quad a_1\geq a_2\geq\cdots\geq a_n \geq\cdots$ .
We will adderess the case of "a increasing sequence", as both cases are same.
If a increasing sequence is bounded above, it has $\sup a_n=A$ (by the real number system property). Therefore for any $\epsilon>0$,
\[ A-\epsilon < a_m\leq a_{m+1}\leq \cdots \leq A \]
Namely for all $n$ such that $n>m$ , $|a_n-A|<\epsilon $ because $A-\epsilon<a_n\leq A$ .
As it means $\lim_{n\rightarrow\infty}a_n=A$ , we get the desired result.
If a real number sequence increasing or decreasing monotonically is bounded, it will converge.
We shall prove it simply. A monotonical sequence is either
$(increasing)\quad a_1\leq a_2\leq\cdots\leq a_n\leq\cdots$ , or
$(decreasing)\quad a_1\geq a_2\geq\cdots\geq a_n \geq\cdots$ .
We will adderess the case of "a increasing sequence", as both cases are same.
If a increasing sequence is bounded above, it has $\sup a_n=A$ (by the real number system property). Therefore for any $\epsilon>0$,
\[ A-\epsilon < a_m\leq a_{m+1}\leq \cdots \leq A \]
Namely for all $n$ such that $n>m$ , $|a_n-A|<\epsilon $ because $A-\epsilon<a_n\leq A$ .
As it means $\lim_{n\rightarrow\infty}a_n=A$ , we get the desired result.