# 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  $f\left(a\right)$ controls it's neighbouring values.

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

 [6.4.1]

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

$f\left(a\right)=\mathrm{lim}f\left({a}_{n}\right)\ge c$

in contrast to  $f\left(a\right).

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:

 $f\left(a\right)\ne 0$ implies  $f\left(x\right)\ne 0$ 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 ${\mathrm{X}}^{2}\left(0\right)\le 0$ but this estimate fails in every neighbourhood of 0.

• The continuity of  f  is essential for [6.4.1] to be valid. If, for $x\in ℝ$,

we have  $f\left(0\right)<0$, 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  $f:A\to ℝ$ and $g:B\to ℝ$ be continuous at $a\in A\cap B$. If  $f\left(a\right) there is a relative $\epsilon$-neighbourhood ${\left(A\cap B\right)}_{a,\epsilon }$ of a such that

 [6.4.3]

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

Secondly we examine whether the neighbouring values have any impact on  $f\left(a\right)$. 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 $A\subset ℝ$ be arbitrary. A real number a is called an accumulation point of A if there is

 a sequence $\left({a}_{n}\right)$ in $A\\left\{a\right\}$ converging to a. [6.4.4]

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

Consider:

• Postulating that $\left({a}_{n}\right)$ is a sequence in $A\\left\{a\right\}$ 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 $\left[0,1\left[\cup \left\{2\right\}$ whereas 2 and 3 are isolated points of this set.

• As $A\\left\{a\right\}=\left(A\\left\{a\right\}\right)\\left\{a\right\}$ we have: a is an accumulation point of A if and only if it is an accumulation point of $A\\left\{a\right\}$.

• Accumulation points of sets and accumulation points of sequences are different. The number 1 for instance is an accumulation point of the sequence $\left({\left(-1\right)}^{n}\right)$. The set of all sequence members $\left\{{\left(-1\right)}^{n}|n\in {ℕ}^{\ast }\right\}=\left\{-1,1\right\}$ 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 ${A}_{a,\epsilon }\subset \left\{a\right\}$.

Example:

 Finite sets do not have accumulation points. [6.4.5]
 All interior points of an interval I are accumulation points of I. [6.4.6]
 The boundary points of an interval I are accumulation points of I. [6.4.7]
 Every real number is an accumulation point of $ℚ$. [6.4.8]

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 $A\\left\{a\right\}=\varnothing$. So we may assume that $A\\left\{a\right\}\ne \varnothing$ and we thus can set

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

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

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

• For $a\in ℚ$ we take $\left(a+\frac{1}{n}\right)$ as a sequence in $ℚ\\left\{a\right\}$ converging to a and we are done. Now assume $a\in ℝ\ℚ$. 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  $\left\{0,1,\dots ,9\right\}$. Thus for each $k\in {ℕ}^{\ast }$ we have:

${b}_{k}≔{x}_{0}\cdot {10}^{n}+\dots +{x}_{n}\cdot {10}^{0}+\sum _{i=1}^{k}\frac{{x}_{i+n}}{{10}^{i}}\in ℚ$

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

Now we can show that the value  $f\left(a\right)$ of a continuous function is controlled by the neighbouring values as soon as a is an accumulation point.

Proposition:  Let  $f:A\to ℝ$ 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 $c\in ℝ$ the following implication holds:

 [6.4.9]

Proof:  As a is an accumulation point of A we find a sequence $\left({a}_{n}\right)$ in $A\\left\{a\right\}$ 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  $f\left({a}_{n}\right)\le c$ and thus, employing [5.5.3] again, we have:  $f\left(a\right)=\mathrm{lim}f\left({a}_{n}\right)\le c$.

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 ${\mathrm{X}}^{2}\left(0\right)=0$.

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

 [6.4.10]

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

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 $ℝ$is solely determined by it's values on $ℚ$. 