6.4. Properties of Continuous Functions
In 5.5 we learnt that limit and sequence members of a convergent sequence mutually control their position. This part will show how this behaviour affects the continuous functions.
At first we find out how the value $f(a)$ controls it's neighbouring values.
Proposition: Let $f:A\to \mathbb{R}$ be continuous at $a\in A$ and let $c\in \mathbb{R}$ be arbitrary. If $f(a)<c$ there is a relative $\epsilon$neighbourhood $A}_{a,\epsilon$ of a such that
$f(x)<c\text{for all}x\in {A}_{a,\epsilon}$

[6.4.1] 
Proof: If there is no such neighbourhood any $n\in {\mathbb{N}}^{\ast}$ we will be assigned a number $a}_{n}\in {A}_{a,\frac{1}{n}$ such that $f({a}_{n})\ge c$. In other words we get a sequence $({a}_{n})$ in A satisfying ${a}_{n}a<\frac{1}{n}$. So we see that ${a}_{n}\to a$. Employing [5.5.3] we thus have
$f(a)=\mathrm{lim}f({a}_{n})\ge c$
in contrast to $f(a)<c$.

Consider:
Of course [6.4.1] is also valid for $>$ and thus holds for $\ne$ as well. Often a special case of [6.4.1] is used:
$f(a)\ne 0$ implies $f(x)\ne 0$ in a suitable neighbourhood $A}_{a,\epsilon$

[6.4.2] 
[6.4.1] is not extendable to $\le$ and $\ge$. The continuous square function for example satisfies ${\mathrm{X}}^{2}(0)\le 0$ but this estimate fails in every neighbourhood of 0.

The continuity of f is essential for [6.4.1] to be valid. If, for $x\in \mathbb{R}$,
$f(x)\u2254\{\begin{array}{l}1\text{, if}x=0\hfill \\ 1\text{, if}x\ne 0\hfill \end{array}$
we have $f(0)<0$, but this is the only negative value of f.
The standard difference trick enables [6.4.1] to be used for the comparison of two functions.
Proposition: Let $f:A\to \mathbb{R}$ and $g:B\to \mathbb{R}$ be continuous at $a\in A\cap B$. If $f(a)<g(a)$ there is a relative $\epsilon$neighbourhood $(A\cap B)}_{a,\epsilon$ of a such that
$f(x)<g(x)\text{for all}x\in {(A\cap B)}_{a,\epsilon}$

[6.4.3] 
Proof: $fg$ is continuous at a and satisfies $fg(a)<0$. From that we get the assertion using [6.4.1].

Secondly we examine whether the neighbouring values have any impact on $f(a)$. Different from the context above however, sufficiently many neighbours are needed for an impact to work. So a must not be isolated but needs to be an accumulation point of A.
Definition: Let $A\subset \mathbb{R}$ be arbitrary. A real number a is called an accumulation point of A if there is
a sequence $({a}_{n})$ in $A\backslash \left\{a\right\}$ converging to a.

[6.4.4] 
a is an isolated point of A if it is no accumulation point of A.

Consider:
Postulating that $({a}_{n})$ is a sequence in $A\backslash \left\{a\right\}$ especially excludes constant sequences. If we were allowed to take such sequences every point would be an accumulation point of any non empty set.
Accumulation points and isolated points may be members of A but they don't need to. For example: 0 and 1 are accumulation points of $[0,1[\cup \left\{2\right\}$ whereas 2 and 3 are isolated points of this set.
As $A\backslash \left\{a\right\}=(A\backslash \{a\})\backslash \left\{a\right\}$ we have: a is an accumulation point of A if and only if it is an accumulation point of $A\backslash \left\{a\right\}$.
Accumulation points of sets and accumulation points of sequences are different. The number 1 for instance is an accumulation point of the sequence $({(1)}^{n})$. The set of all sequence members $\{{(1)}^{n}n\in {\mathbb{N}}^{\ast}\}=\{\mathrm{1,1}\}$ however has no accumulation points at all.
a is an accumulation point of A if and only if each $\epsilon$neighbourhood of a contains at least one member of A different from a.
Therefor a is isolated if and only if there is a relative $\epsilon$neighbourhood $A}_{a,\epsilon$ such that $A}_{a,\epsilon}\subset \{a\$.
Example:
Proof:

Let a be arbitrary and let A be finite. If $A\subset \left\{a\right\}$ a certainly fails to be an accumulation point of A because there are no sequences in $A\backslash \left\{a\right\}=\varnothing$. So we may assume that $A\backslash \left\{a\right\}\ne \varnothing$ and we thus can set
$\epsilon \u2254\mathrm{min}\{xax\in A\backslash \left\{a\right\}\}$
For this special $\epsilon$ we know that $A}_{a,\epsilon$ contains no member of A different from a. Thus a is isolated.

If a is an interior point of I the members of $(a+\frac{1}{n})$ are all inside I from a certain $n}_{0$ onwards. The sequence $(a+\frac{1}{n+{n}_{0}})$ thus proves a to be an accumulation point of I.

For a left boundary point a of I we again employ the sequence $(a+\frac{1}{n})$ 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 $(a\frac{1}{n})$.

For $a\in \mathbb{Q}$ we take $(a+\frac{1}{n})$ as a sequence in $\mathbb{Q}\backslash \left\{a\right\}$ converging to a and we are done. Now assume $a\in \mathbb{R}\backslash \mathbb{Q}$. According to [5.9.5] the number $a$ has a decimal representation
$a={x}_{0}\cdot {10}^{n}+\dots +{x}_{n}\cdot {10}^{0}+\sum _{i=1}^{\infty}\frac{{x}_{i+n}}{{10}^{i}}$
with the decimal places $x}_{j$ being members of $\{\mathrm{0,1,}\dots \mathrm{,9}\}$. Thus for each $k\in {\mathbb{N}}^{\ast}$ we have:
$b}_{k}\u2254{x}_{0}\cdot {10}^{n}+\dots +{x}_{n}\cdot {10}^{0}+\sum _{i=1}^{k}\frac{{x}_{i+n}}{{10}^{i}}\in \mathbb{Q$
so that $({b}_{n})$ is a sequence in $\mathbb{Q}$ converging to $a$. This proves the assertion if $a\ge 0$. In case $a<0$ we get the result with the sequence $({b}_{n})$.

Now we can show that the value $f(a)$ of a continuous function is controlled by the neighbouring values as soon as a is an accumulation point.
Proposition: Let $f:A\to \mathbb{R}$ be continuous at $a\in A$ and let a be an accumulation point of A. Then for each relative $\epsilon$neighbourhood $A}_{a,\epsilon$ and every $c\in \mathbb{R}$ the following implication holds:
$f(x)\le c\text{for all}x\in {A}_{a,\epsilon}\backslash \{a\}\text{\hspace{1em}}\Rightarrow \text{\hspace{1em}}f(a)\le c$

[6.4.9] 
Proof: As a is an accumulation point of A we find a sequence $({a}_{n})$ in $A\backslash \left\{a\right\}$ converging to a. So there is an $n}_{0$ such that $a}_{n}\in {A}_{a,\epsilon$ for all $n\ge {n}_{0}$. Due to the premise these n satisfy $f({a}_{n})\le c$ and thus, employing [5.5.3] again, we have: $f(a)=\mathrm{lim}f({a}_{n})\le c$.

Consider:
[6.4.9] is also provable for $\ge$ (and thus for $=$).
[6.4.9] is not restrictable to $<$ or to $>$. The values of the continuous square function for example are strictly positive except for ${\mathrm{X}}^{2}(0)=0$.
Also, the continuity of f at a as well as the accumulation property of a are indispensable premises.
[6.4.9] can be used parallel to [6.4.3] if comparing two functions. As an example we note this for the equality relation.
Proposition: Let $f:A\to \mathbb{R}$ and $g:B\to \mathbb{R}$ be continuous at $a\in A\cap B$ and a be an accumulation point of $A\cap B$. Then every relative $\epsilon$neighbourhood $(A\cap B)}_{a,\epsilon$ satisfies:
$f(x)=g(x)\text{for all}x\in {(A\cap B)}_{a,\epsilon}\backslash \{a\}\text{\hspace{1em}}\Rightarrow \text{\hspace{1em}}f(a)=g(a)$

[6.4.10] 
Proof: As $fg$ is continuous at a and as $fg(x)=0$ for all $x\in {(A\cap B)}_{a,\epsilon}\backslash \{a\}$ we conclude that $fg(a)=0$.

So we see that with continuous functions the values at accumulation points of their domain are uniquely determined by the neighbouring values. According to [6.4.8] for instance any continuous function on $\mathbb{R}$is solely determined by it's values on $\mathbb{Q}$.
