6.4. Properties of Continuous Functions

Proposition:  Let  ${\displaystyle f:A\to ℝ}$ be continuous at ${\displaystyle a\in A}$ and let ${\displaystyle c\in ℝ}$ be arbitrary. If  ${\displaystyle f(a)<c}$ there is a relative ${\displaystyle \epsilon }$-neighbourhood ${\displaystyle A_{a,\epsilon }}$ of a such that

Proof:  If there is no such neighbourhood any ${\displaystyle n\in {\mathbb{N}}^{\ast }}$ we will be assigned a number ${\displaystyle a_n\in A_{a,\frac{1}{n}}}$ such that  ${\displaystyle f(a_n)\ge c}$. In other words we get a sequence $(a_n)$ in A satisfying ${\displaystyle |a_n-a|<\frac{1}{n}}$. So we see that ${\displaystyle a_n\to a}$. Employing [5.5.3] we thus have

in contrast to  ${\displaystyle f(a)<c}$.

Proposition:  Let  ${\displaystyle f:A\to ℝ}$ and ${\displaystyle g:B\to ℝ}$ be continuous at ${\displaystyle a\in A\cap B}$. If  ${\displaystyle f(a)<g(a)}$ there is a relative ${\displaystyle \epsilon }$-neighbourhood ${\displaystyle {(A\cap B)}_{a,\epsilon }}$ of a such that

Proof:  ${\displaystyle f-g}$ is continuous at a and satisfies  ${\displaystyle f-g(a)<0}$. From that we get the assertion using [6.4.1].

Definition:  Let ${\displaystyle A\subset ℝ}$ be arbitrary. A real number a is called an accumulation point of A if there is

a is an isolated point of A if it is no accumulation point of A.

Example:  

Proof:  

• Let a be arbitrary and let A be finite. If ${\displaystyle A\subset \left\{a\right\}}$  a certainly fails to be an accumulation point of A because there are no sequences in ${\displaystyle A\backslash \left\{a\right\}=\varnothing }$. So we may assume that ${\displaystyle A\backslash \left\{a\right\}\ne \varnothing }$ and we thus can set

  ${\displaystyle \epsilon ≔\min \{|x-a||x\in A\backslash \left\{a\right\}\}}$

  For this special ${\displaystyle \epsilon }$ we know that ${\displaystyle A_{a,\epsilon }}$ contains no member of A different from a. Thus a is isolated.

• If a is an interior point of I the members of ${\displaystyle (a+\frac{1}{n})}$ are all inside I from a certain ${\displaystyle n_0}$ onwards. The sequence ${\displaystyle (a+\frac{1}{n+n_0})}$ thus proves a to be an accumulation point of I.

• For a left boundary point a of I we again employ the sequence ${\displaystyle (a+\frac{1}{n})}$ and proceed the same way as above. The proof for a right boundary point a is quite similar. We just need to substitute the starting sequence by ${\displaystyle (a-\frac{1}{n})}$.

• For ${\displaystyle a\in \mathbb{Q}}$ we take ${\displaystyle (a+\frac{1}{n})}$ as a sequence in ${\displaystyle \mathbb{Q}\backslash \left\{a\right\}}$ converging to a and we are done. Now assume ${\displaystyle a\in ℝ\backslash \mathbb{Q}}$. According to [5.9.5] the number ${\displaystyle |a|}$ has a decimal representation

  ${\displaystyle |a|=x_0\cdot {10}^n+\dots +x_n\cdot {10}^0+\sum _{i=1}^{\infty }\frac{x_{i+n}}{{10}^i}}$

  with the decimal places ${\displaystyle x_j}$ being members of  ${\displaystyle \{\mathrm{0,1,}\dots \mathrm{,9}\}}$. Thus for each ${\displaystyle k\in {\mathbb{N}}^{\ast }}$ we have:

  ${\displaystyle b_k≔x_0\cdot {10}^n+\dots +x_n\cdot {10}^0+\sum _{i=1}^k\frac{x_{i+n}}{{10}^i}\in \mathbb{Q}}$

  so that ${\displaystyle (b_n)}$ is a sequence in ${\displaystyle \mathbb{Q}}$ converging to ${\displaystyle |a|}$. This proves the assertion if ${\displaystyle a\ge 0}$. In case ${\displaystyle a<0}$ we get the result with the sequence  ${\displaystyle (-b_n)}$.

Proposition:  Let  ${\displaystyle f:A\to ℝ}$ be continuous at ${\displaystyle a\in A}$ and let a be an accumulation point of A. Then for each relative ${\displaystyle \epsilon }$-neighbourhood ${\displaystyle A_{a,\epsilon }}$ and every ${\displaystyle c\in ℝ}$ the following implication holds:

Proof:  As a is an accumulation point of A we find a sequence $(a_n)$ in ${\displaystyle A\backslash \left\{a\right\}}$ converging to a. So there is an ${\displaystyle n_0}$ such that ${\displaystyle a_n\in A_{a,\epsilon }}$ for all ${\displaystyle n\ge n_0}$. Due to the premise these n satisfy  ${\displaystyle f(a_n)\le c}$ and thus, employing [5.5.3] again, we have:  ${\displaystyle f(a)=\lim f(a_n)\le c}$.

Proposition:  Let  ${\displaystyle f:A\to ℝ}$ and ${\displaystyle g:B\to ℝ}$ be continuous at ${\displaystyle a\in A\cap B}$ and a be an accumulation point of ${\displaystyle A\cap B}$. Then every relative ${\displaystyle \epsilon }$-neighbourhood ${\displaystyle {(A\cap B)}_{a,\epsilon }}$ satisfies:

Proof:  As  ${\displaystyle f-g}$ is continuous at a and as  ${\displaystyle f-g(x)=0}$ for all ${\displaystyle x\in {(A\cap B)}_{a,\epsilon }\backslash \{a\}}$ we conclude that  ${\displaystyle f-g(a)=0}$.