# 6.3. Calculation Rules for Continuous Functions

Basically the concept of continuity is the concept of convergence. So it comes with no surprise that a number of properties of convergent sequences are mirrored in the properties of continuous functions. At first in this part we transfer the limit theorems. Furthermore we find that composition is compatible with continuity as well.

Proposition (Local Continuity Theorems):  Let  $f:A\to ℝ$ and $g:B\to ℝ$ be any two functions and $n\in {ℕ}^{\ast }$ be arbitrary. If  f and g are continuous at $a\in A\cap B$ so is

 $f+g$ [6.3.1]
 $f-g$ [6.3.2]
 $f\cdot g$ and thus  ${f}^{n}$ [6.3.3]
 $\frac{f}{g}$ and especially $\frac{1}{g}$ provided that $g\left(a\right)\ne 0$ [6.3.4]

If g is continuous at a and if  f is continuous at $g\left(a\right)\in A$ then

 $f\circ g$ is continuous at a [6.3.5]

Proof:  The first four assertions are easily traced back to the respective limit theorem. As an example we show this for the first theorem.

Let $\left({a}_{n}\right)$ be a sequence in $A\cap B$ converging to a. As $A\cap B\subset A$ and $A\cap B\subset B$ we can employ the continuity of  f and g to get:

• $A\ni {a}_{n}\to a\text{ }⇒\text{ }f\left({a}_{n}\right)\to f\left(a\right)$
• $B\ni {a}_{n}\to a\text{ }⇒\text{ }g\left({a}_{n}\right)\to g\left(a\right)$

Thus the first limit theorem assures that

$f+g\left({a}_{n}\right)=f\left({a}_{n}\right)+g\left({a}_{n}\right)\to f\left(a\right)+g\left(a\right)=f+g\left(a\right)$.

To prove the fifth theorem we take any sequence $\left({a}_{n}\right)$ in $\left\{x\in B|g\left(x\right)\in A\right\}\subset B$ converging to a. As $g\left({a}_{n}\right)\in A$ the following two step conclusion holds due to the premises:

${a}_{n}\to a\text{ }⇒\text{ }g\left({a}_{n}\right)\to g\left(a\right)\text{ }⇒\text{ }f\circ g\left({a}_{n}\right)=f\left(g\left({a}_{n}\right)\right)\to f\left(g\left(a\right)\right)=f\circ g\left(a\right)$

Having proved the pointwise version of the continuity theorems we change to a global view. In the following proposition we just restate the continuity theorems according to this idea.

Proposition (Global Continuity Theorems):

 $f\in {\mathcal{C}}^{0}\left(A\right)\text{\hspace{0.28em}}\wedge \text{\hspace{0.28em}}g\in {\mathcal{C}}^{0}\left(B\right)\text{ }⇒\text{ }f+g\in {\mathcal{C}}^{0}\left(A\cap B\right)$ [6.3.6]
 $f\in {\mathcal{C}}^{0}\left(A\right)\text{\hspace{0.28em}}\wedge \text{\hspace{0.28em}}g\in {\mathcal{C}}^{0}\left(B\right)\text{ }⇒\text{ }f-g\in {\mathcal{C}}^{0}\left(A\cap B\right)$ [6.3.7]
 $f\in {\mathcal{C}}^{0}\left(A\right)\text{\hspace{0.28em}}\wedge \text{\hspace{0.28em}}g\in {\mathcal{C}}^{0}\left(B\right)\text{ }⇒\text{ }f\cdot g\in {\mathcal{C}}^{0}\left(A\cap B\right)$ [6.3.8]
 $f\in {\mathcal{C}}^{0}\left(A\right)\text{\hspace{0.28em}}\wedge \text{\hspace{0.28em}}g\in {\mathcal{C}}^{0}\left(B\right)\text{ }⇒\text{ }\frac{f}{g}\in {\mathcal{C}}^{0}\left(\left\{x\in A\cap B|g\left(x\right)\ne 0\right\}\right)$ [6.3.9]
 $f\in {\mathcal{C}}^{0}\left(A\right)\text{\hspace{0.28em}}\wedge \text{\hspace{0.28em}}g\in {\mathcal{C}}^{0}\left(B\right)\text{ }⇒\text{ }f\circ g\in {\mathcal{C}}^{0}\left(\left\{x\in B|g\left(x\right)\in A\right\}\right)$ [6.3.10]

Consider:

• Calculation rules nearly always allow to provide the domain of discourse, the function space ${\mathcal{C}}^{0}\left(A\right)$ in this case, with an algebraic structure. As the constant functions $\mathbf{0}≔0|A$ and $\mathbf{1}≔1|A$ are continuous we get with [6.3.6] to [6.3.8]:

$\left({\mathcal{C}}^{0}\left(A\right),+\right)$ is an abelian group  i The addition + is associative and commutative. 0 is the neutral element, i.e.  $f+\mathbf{0}=f$  for all  f. Each f has a unique inverse, $-f$ in this case, such that  $f+\left(-f\right)=\mathbf{0}$.
.
$\left({\mathcal{C}}^{0}\left(A\right),+,\cdot \right)$ is a commutative ring with identity element  i The axioms for an abelian group are satisfied. The multiplication · is associative and commutative. · is distributive with respect to +. 1 is the neutral element for the multiplication, i.e. $\mathbf{1}·f=f$ for all f.
.
• The continuity of the polynomials and their quotients (already proved in [6.2.17]) could be regained from [6.3.6] and [6.3.8] and [6.3.9] respectively without knowing anything about analytical functions.

• To prove the continuity of the trigonometric functions ([6.2.17]) more simply we only need to know that sin is continuous. Because then we would know the same for

$\mathrm{cos}=\mathrm{sin}\circ \left(\mathrm{X}+\frac{\pi }{2}\right)$  from [6.3.10]

$\mathrm{tan}=\frac{\mathrm{sin}}{\mathrm{cos}}$  and  $\mathrm{cot}=\frac{\mathrm{cos}}{\mathrm{sin}}$  from [6.3.9]

The set $\mathcal{Bi}\left(A\right)$ of bijections from A to A is a (non abelian) group under the composition $\circ$ with $\mathrm{X}|A$ as its identity element. According to [6.3.10]  $\circ$ also operates on $\mathcal{Bi}\left(A\right)\cap {\mathcal{C}}^{0}\left(A\right)$. The question whether this is a group operation is solely the question whether the inverse of a continuous function is continuous again. Surprisingly the answer is only a matter of the domain.

Proposition:  For any bijection  $f:I\to B\subset ℝ$ on an interval
 i We understand this as a common notation for open and closed intervals. In the open case we also allow the values $\infty$ for the right and $-\infty$ for the left boundery. Thus $ℝ$ and e.g. ${ℝ}^{>0}$ are regarded as intervals as well.
we have:

 f  continuous$\text{ }⇒\text{ }{f}^{-1}$ continuous [6.3.11]

Proof:  We start with a closed interval I and proceed indirectly. Suppose that there is an $a\in B$ and a sequence $\left({a}_{n}\right)$ in B such that ${a}_{n}\to a$ but with $\left({f}^{-1}\left({a}_{n}\right)\right)$ not converging to  ${f}^{-1}\left(a\right)$. Then we find an $\epsilon >0$ such that the special $\epsilon$-neighbourhood $\right]$ lacks almost all of the sequence members. Thus we can create the recursion

$\begin{array}{l}{k}_{1}>1\text{ }\wedge \text{ }{f}^{-1}\left({a}_{{k}_{1}}\right)\notin \right]{f}^{-1}\left(a\right)-\epsilon ,{f}^{-1}\left(a\right)+\epsilon \left[\\ {k}_{n+1}>{k}_{n}\text{ }\wedge \text{ }{f}^{-1}\left({a}_{{k}_{n+1}}\right)\notin \right]{f}^{-1}\left(a\right)-\epsilon ,{f}^{-1}\left(a\right)+\epsilon \left[\end{array}$

to get a subsequence $\left({f}^{-1}\left({a}_{{k}_{n}}\right)\right)$ in I with all it's members being outside $\right]{f}^{-1}\left(a\right)-\epsilon ,{f}^{-1}\left(a\right)+\epsilon \left[$ [1]. As I is closed $\left({f}^{-1}\left({a}_{{k}_{n}}\right)\right)$ is a bounded sequence and thus has an accumulation point b according to the Bolzano-Weierstrass theorem [5.8.6]. Due to [5.8.5] there is now a further subsequence $\left({f}^{-1}\left({a}_{{k}_{{j}_{n}}}\right)\right)$ such that

${f}^{-1}\left({a}_{{k}_{{j}_{n}}}\right)\to b$

With I being closed we have $b\in I$ and as  f  is continuous at b we see ${a}_{{k}_{{j}_{n}}}\to f\left(b\right)$. On the other hand we know that ${a}_{{k}_{{j}_{n}}}\to a$. Therefor we have  $f\left(b\right)=a$ i.e. $b={f}^{-1}\left(a\right)$. But this actually means ${f}^{-1}\left({a}_{{k}_{{j}_{n}}}\right)\to {f}^{-1}\left(a\right)$  and thus contradicts [1].

Now let I be any interval and $a\in B$ be arbitrary. We find a closed interval $J\subset I$ with ${f}^{-1}\left(a\right)\in J$. The restriction  $f|J$ is continuous on J and using a result from part 6 ([6.6.6]) we find that  $f\left(J\right)$ is an interval. Thus  ${f}^{-1}$ and ${\left(f|J\right)}^{-1}$ locally coincide at a. So  ${f}^{-1}$ is continuous at a because we know from the first part that ${\left(f|J\right)}^{-1}$ is continuous at a.

Consider:

• A bijection that is continuous bothways is called a homoeomorphism. Thus [6.3.11] says:

Every continuous bijection on an interval is a homoeomorphism.

• As $\mathrm{X}|I$ is continuous we now have for each interval I :

$\left(\mathcal{Bi}\left(I\right)\cap {\mathcal{C}}^{0}\left(I\right),\circ \right)$ is a non abelian group.

• Generally [6.3.11] no longer holds if the domain of  f  is no interval. Take for example the function  $f:\right]0,1\right]\cup \left\{2\right\}\to \left[0,1\right]$ given by

f  is continuous (at 2, as 2 is isolated. At $x\in \right]0,1\right]$  f  locally coincides with X) and is bijective:

${f}^{-1}$ however is discontinuous at 0.

 6.2. 6.4.