6.2. Continuous Functions
Convergence occupies a central position among the properties of sequences. It is thus inevitable to study how functions deal with convergent sequences. We will distinguish those functions that are convergencepreserving, i.e. those that hand down convergence to the image sequences.
Definition: Let $a\in A\subset \mathbb{R}$. A function $f:A\to \mathbb{R}$ is called continuous at a, if the following holds for every sequence $({a}_{n})$ in A:
${a}_{n}\to a\text{\hspace{1em}}\Rightarrow \text{\hspace{1em}}f({a}_{n})\to f(a)$

[6.2.1] 

Note how this definition elaborates the property 'convergencepreserving': It isn't just simply convergence that we claim for an image sequence $(f({a}_{n}))$ but convergence to the 'right' limit $f(a)$. From that we see that the value $f(a)$ is related in certain way to it's neighbouring values: No matter how we approach a on the xaxis, (i.e. whichever sequence ${a}_{n}\to a$ we choose), the image sequence on the yaxis will converge to $f(a)$.
Consider:

There are two ways to to show that a function f is discontinuous at a:
to find a sequence $({a}_{n})$ in A converging to a such that $(f({a}_{n}))$ is divergent.
to find a sequence $({a}_{n})$ in A converging to a such that $(f({a}_{n}))$ converges to a number different from $f(a)$.

The more handy notion of continuity
$\mathrm{lim}\text{\hspace{0.17em}}f({a}_{n})=f(\mathrm{lim}\text{\hspace{0.17em}}{a}_{n})$,

[6.2.2] 
reveals the following aspect: Continuous functions allow to swap the order of calculation of the limit and evaluating the function. This opens advanced techniques to calculate limits.
Every function f is continuous at a if a is an isolated point of A. This is because any sequence $({a}_{n})$ in A converging to a will turn out to be almost constant. Consequently this applies to the image sequence as well so that the convergence $f({a}_{n})\to f(a)$ is valid.
The first series of examples covers some common functions. All of them are continuous at every point of their domain. The first four examples are easy to prove using the limit theorems.
Example:

[6.2.3] 
because: $\mathbb{R}\ni {a}_{n}\to a\text{\hspace{1em}}\Rightarrow \text{\hspace{1em}}c({a}_{n})=c\to c=c(a)$ 

[6.2.4] 
because: $\mathbb{R}\ni {a}_{n}\to a\text{\hspace{1em}}\Rightarrow \text{\hspace{1em}}m\mathrm{X}+b({a}_{n})=m{a}_{n}+b\to ma+b=m\mathrm{X}+b(a)$ 

[6.2.5] 
because: $\mathbb{R}\ni {a}_{n}\to a\text{\hspace{1em}}\Rightarrow \text{\hspace{1em}}{\mathrm{X}}^{k}({a}_{n})={a}_{n}^{\text{\hspace{0.05em}}k}\to {a}^{k}={\mathrm{X}}^{k}(a)$ 

[6.2.6] 
because: $0\ne {a}_{n}\to a\text{\hspace{1em}}\Rightarrow \text{\hspace{1em}}\frac{1}{{\mathrm{X}}^{k}}({a}_{n})=\frac{1}{{a}_{n}^{\text{\hspace{0.05em}}k}}\to \frac{1}{{a}^{k}}=\frac{1}{{\mathrm{X}}^{k}}(a)$


[6.2.7] 
Let $({a}_{n})$ be a sequence in ${\mathbb{R}}^{\ge 0}$ such that ${a}_{n}\to a$. We have to show: $\sqrt{{a}_{n}}\to \sqrt{a}$. To that end we take an $\epsilon >0$ and proceed separately according to the value of a:
 $a=0$: For ${\epsilon}^{2}$ there is an ${n}_{0}$ such that ${a}_{n}={a}_{n}0<{\epsilon}^{2}$ for all $n\ge {n}_{0}$. These n satisfy:
$\sqrt{{a}_{n}}\sqrt{0}=\sqrt{{a}_{n}}<\sqrt{{\epsilon}^{2}}=\epsilon$
 $a>0$: For $\epsilon \sqrt{a}>0$ there is an $n}_{0$ such that ${a}_{n}a<\epsilon \sqrt{a}$ for all $n\ge {n}_{0}$. With these n we can estimate as follows:
$\sqrt{{a}_{n}}\sqrt{a}=\frac{\sqrt{{a}_{n}}\sqrt{a}\cdot \sqrt{{a}_{n}}+\sqrt{a}}{\sqrt{{a}_{n}}+\sqrt{a}}=\frac{{a}_{n}a}{\sqrt{{a}_{n}}+\sqrt{a}}\le \frac{{a}_{n}a}{\sqrt{a}}<\frac{\epsilon \sqrt{a}}{\sqrt{a}}=\epsilon$



As already mentioned briefly [6.2.2] enables us to cope with some more challenging sequences. Using the root function's continuity for example we easily see that
 $\sqrt{\frac{4{n}^{3}+3n}{9{n}^{3}2}}\to \sqrt{\frac{4}{9}}=\frac{2}{3}$
 $\sqrt{3+\sqrt{\frac{n1}{n+1}}}\to \sqrt{3+\sqrt{1}}=2$
Being closely connected with the limit concept continuity ata distinguishes such functions that have certain properties in the vicinity of a. Continuity is thus a local property!
We will show aspects of this property in detail within the next two propositions. At first we prove that cutting down the domain will not effect the continuity at a.
Proposition: For any function $f:B\to \mathbb{R}$ and arbitrary $a\in A\subset B$ we have
f is continuous at a$\text{\hspace{1em}}\Rightarrow \text{\hspace{1em}}fA$ is continuous at a

[6.2.8] 
In general the reverse is false: According to example [6.2.13] the
Heaviside step function H
i 
$\mathrm{H}(x)=\{\begin{array}{l}1\text{, if}x\ge 0\hfill \\ 0\text{, if}x0\hfill \end{array}$

is discontinuous at 0, althought it's restriction $\mathrm{H}{\mathbb{R}}^{\ge 0}=1{\mathbb{R}}^{\ge 0}$ is continuous at this point.
Proof: As $A\subset B$ we find that every sequence $({a}_{n})$ in A converging to a is also a sequence in B with limit a. From that we see:
$fA({a}_{n})=f({a}_{n})\to f(a)=fA(a)$

In a second example we prove that the continuity behaviour of two functions is identical as soon as they coincide in a neighbourhood of a. To state this precisely we introduce the following notions.
For any $a\in \mathbb{R}$ and $\epsilon >0$ the set
${A}_{a,\epsilon}\u2254A\cap ]a\epsilon ,a+\epsilon [$

[6.2.9] 
is called a relative εneighbourhood of a.
Two functions $f:A\to \mathbb{R}$ and $g:B\to \mathbb{R}$ are said to coincide locally at a if there is an $\epsilon >0$ such that $A}_{a,\epsilon}={B}_{a,\epsilon$ and
$f(x)=g(x)\text{for all}x\in {A}_{a,\epsilon}={B}_{a,\epsilon}$

[6.2.10] 
Proposition: If $f:A\to \mathbb{R}$ and $g:B\to \mathbb{R}$ coincide locally at $a\in A\cap B$ we have
f is continuous at a$\text{\hspace{1em}}\iff \text{\hspace{1em}}$g is continuous at a

[6.2.11] 
Proof: We only show "$\Rightarrow$" as we get the other direction simply by interchanging f and g. Now if $({a}_{n})$ is an arbitrary sequence in B converging to a we find a suitable $n}_{0$ so that all sequence members are to be found in $B}_{a,\epsilon}={A}_{a,\epsilon$
. Taking $n\ge {n}_{0}$ we thus conclude:
$g({a}_{n})=f({a}_{n})\to f(a)=g(a)$

Consider:
Obviously the functions f and $f{A}_{a,\epsilon}$ coincide locally at a for each $\epsilon >0$. Thus we get the following equivalence as a special case of [6.2.11]:
f is continuous at a$\text{\hspace{1em}}\iff \text{\hspace{1em}}f{A}_{a,\epsilon}$ is continuous at a for one $\epsilon >0$
Piecewise defined functions often benefit from [6.2.11]. As an example we take the absolute value function and the Heaviside step function:
Example:

[6.2.12] 
At $a>0$ $\mathrm{X}$ locally coincides with the linear function X and thus is continuous at a.
At $a<0$ $\mathrm{X}$ locally coincides with the linear function $\mathrm{X}$ and thus is continuous at a.
The case $a=0$ cannot be solved by the 'locally coincide' trick. But we will succeed with a
result on zero sequences
i 
${a}_{n}\to 0\text{\hspace{1em}}\Rightarrow \text{\hspace{1em}}{a}_{n}\to 0$

( [5.5.6] ) instead:
${a}_{n}\to 0\text{\hspace{1em}}\Rightarrow \text{\hspace{1em}}\mathrm{X}({a}_{n})={a}_{n}\to 0=\mathrm{X}(a)$


[6.2.13] 
At $a>0$ H locally coincides with the constant function 1. Therefor H is continuous at a.
At $a<0$ H locally coincides with the constant function 0 and thus is continuous at a.
An
example
i 
$\frac{{(1)}^{n}}{n}\to 0\text{, but}(\mathrm{H}(\frac{{(1)}^{n}}{n}))\text{is divergent}$

in [6.1] shows that H is discontinuous at 0.


So far we only know functions that are continuous everywhere or everywhere except for one point. The next example will present functions that are continuous nowhere or continuous at only one point. The second example thereby again shows the local character of continuity: The continuity at a will not necessarily be passed on to neighbouring points.
Example:

The indicator function $\mathrm{\chi}}_{\mathbb{Q}$
i 
$\mathrm{\chi}}_{\mathbb{Q}}(x)=\{\begin{array}{l}1\text{, if}x\in \mathbb{Q}\hfill \\ 0\text{, if}x\notin \mathbb{Q}\hfill \end{array$ 
is nowhere continuous:

[6.2.14] 
If $a\in \mathbb{Q}$ take a sequence $({a}_{n})$ in $\mathbb{R}\backslash \mathbb{Q}\subset \mathbb{R}$ converging to a. Then:
${\mathrm{\chi}}_{\mathbb{Q}}({a}_{n})=0\to 0\ne 1={\mathrm{\chi}}_{\mathbb{Q}}(a)$
If $a\in \mathbb{R}\backslash \mathbb{Q}$ take a sequence $({a}_{n})$ in $\mathbb{Q}\subset \mathbb{R}$ converging to a. This time we have:
${\mathrm{\chi}}_{\mathbb{Q}}({a}_{n})=1\to 1\ne 0={\mathrm{\chi}}_{\mathbb{Q}}(a)$


[6.2.15] 

To show the continuity at 0 we take an arbitrary sequence ${a}_{n}\to 0$. As $\mathrm{\chi}}_{\mathbb{Q}}({a}_{n})\in \{\mathrm{0,1}\$ the sequence $(\mathrm{X}\cdot {\mathrm{\chi}}_{\mathbb{Q}}({a}_{n}))=({a}_{n})\cdot ({\mathrm{\chi}}_{\mathbb{Q}}({a}_{n}))$ can be factorized into a zero sequence times a bounded sequence, and thus converges to $0=\mathrm{X}\cdot {\mathrm{\chi}}_{\mathbb{Q}}(0)$.

Now consider $a\ne 0$ and assume that $\mathrm{X}\cdot {\mathrm{\chi}}_{\mathbb{Q}}$ is continuous at a. Then the implication
${a}_{n}\to a\text{\hspace{1em}}\Rightarrow \text{\hspace{1em}}{a}_{n}\cdot {\mathrm{\chi}}_{\mathbb{Q}}({a}_{n})\to a\cdot {\mathrm{\chi}}_{\mathbb{Q}}(a)$
is always true. Due to the fourth limit theorem we then have (without restriction we may assume ${a}_{n}\ne 0$):
${\mathrm{\chi}}_{\mathbb{Q}}({a}_{n})=\frac{{a}_{n}\cdot {\mathrm{\chi}}_{\mathbb{Q}}({a}_{n})}{{a}_{n}}\to \frac{a\cdot {\mathrm{\chi}}_{\mathbb{Q}}(a)}{a}={\mathrm{\chi}}_{\mathbb{Q}}(a)$.
Thus we have proved $\mathrm{\chi}}_{\mathbb{Q}$ to be continuous at a. Contradiction!


Many functions from our examples are continuous at more than one point, often even at each point of their domain. This observation motivates the introduction of a broader concept of continuity.
Definition: Let $A\subset B\subset \mathbb{R}$. A function $f:B\to \mathbb{R}$ is called continuous on A, if f is continuous at each $a\in A$.
We omit the extension 'on A' if $A=B$, i.e. if f is continuous at every point of it's domain. The set of all functions being continuous on A is denoted by the symbol

We restate our examples using the new notation:
 $c\in {\mathcal{C}}^{0}(\mathbb{R}),\text{\hspace{1em}}m\mathrm{X}+b\in {\mathcal{C}}^{0}(\mathbb{R}),\text{\hspace{1em}}{\mathrm{X}}^{k}\in {\mathcal{C}}^{0}(\mathbb{R}),\text{\hspace{1em}}\frac{1}{{\mathrm{X}}^{k}}\in {\mathcal{C}}^{0}({\mathbb{R}}^{\ne 0}),\text{\hspace{1em}}\sqrt{\mathrm{X}}\in {\mathcal{C}}^{0}({\mathbb{R}}^{\ge 0})$
 $\mathrm{X}\in {\mathcal{C}}^{0}(\mathbb{R}),\text{\hspace{1em}}\mathrm{H}\in {\mathcal{C}}^{0}({\mathbb{R}}^{\ne 0}),\text{\hspace{1em}}\mathrm{H}\notin {\mathcal{C}}^{0}(\mathbb{R})$
 ${\mathrm{\chi}}_{\mathbb{Q}}\in {\mathcal{C}}^{0}(\varnothing ),\text{\hspace{1em}}\mathrm{X}\cdot {\mathrm{\chi}}_{\mathbb{Q}}\in {\mathcal{C}}^{0}(\{0\})$
The next part will reveal important algebraic properties of ${\mathcal{C}}^{0}(A)$. In this context we will also examine how continuity responds to basic arithmetics. We expect good results as this strategy has been very successful with convergence.
The remainder of this section is addressed to prove the continuity of the analytical functions. From this and [5.12.4] the continuity of the polynomials and their quotients, of sin, cos, tan, cot and exp is guaranteed.
$p\in {\mathcal{C}}^{0}(\mathbb{R})$, $\frac{p}{q}\in {\mathcal{C}}^{0}(\{x\in \mathbb{R}q(x)\ne 0\})$
$\mathrm{sin},\text{\hspace{0.17em}}\mathrm{cos},\text{\hspace{0.17em}}\mathrm{exp}\in {\mathcal{C}}^{0}(\mathbb{R})$
$\mathrm{tan}\in {\mathcal{C}}^{0}(\{x\in \mathbb{R}\mathrm{cos}x\ne 0\})$, $\mathrm{cot}\in {\mathcal{C}}^{0}(\{x\in \mathbb{R}\mathrm{sin}x\ne 0\})$

[6.2.17] 
Proposition: The limit function of a convergent power series $(\sum _{i=0}^{n}{a}_{i}{(\mathrm{X}a)}^{i})$ is continuous at every point b of it's domain of convergence:
$\sum _{i=0}^{\infty}{a}_{i}{(\mathrm{X}a)}^{i}\in {\mathcal{C}}^{0}(]ar,a+r[)$

[6.2.18] 
Proof: Let $b\in ]ar,a+r[$ be arbitrary. The rearrangement theorem [5.11.20] provides a convergent power series
$(\sum _{i=0}^{n}{b}_{i}{(\mathrm{X}b)}^{i})$ such that
$\sum _{i=0}^{\infty}{b}_{i}{(xb)}^{i}=\sum _{i=0}^{\infty}{a}_{i}{(xa)}^{i}\text{for all}x\text{satisfying}xbs\u2254rba$
The limit functions thus coincide locally at b. So it is sufficient (cf. [6.2.11]) to prove the continuity of $(\sum _{i=0}^{\infty}{b}_{i}{(\mathrm{X}b)}^{i})$ in b. To that end we take an arbitrary convergence $]bs,b+s[\ni {x}_{n}\to b$ and have to show:
$\sum _{i=0}^{\infty}{b}_{i}{({x}_{n}b)}^{i}\to \sum _{i=0}^{\infty}{b}_{i}{(bb)}^{i}={b}_{0}$[1]
At first we get a number $t>0$ such that ${x}_{n}b\le t<s$ for all $n\in {\mathbb{N}}^{\ast}$. According to [5.11.10] the series $(\sum _{i=0}^{n}{b}_{i}{t}^{i})$ is absolutely convergent. Consequently the same is true for $(\sum _{i=1}^{n}{b}_{i}{t}^{i1})$ so that we can estimate as follows:
$0\le \sum _{i=0}^{\infty}{b}_{i}{({x}_{n}b)}^{i}{b}_{0}=({x}_{n}b)\sum _{i=1}^{\infty}{b}_{i}{({x}_{n}b)}^{i1}\le {x}_{n}b\sum _{i=1}^{\infty}{b}_{i}{t}^{i1}$
As ${x}_{n}\to b$ the right hand side converges to 0. This ensures $\sum _{i=0}^{\infty}{b}_{i}{({x}_{n}b)}^{i}{b}_{0}\to 0$ and thus [1] is valid.

By definition an analytical function coincides locally at every point of it's domain with the limit function of a convergent power series. So we know:
Proposition: Every analytical function is continuous, i.e.:
${\mathcal{C}}^{\ast}(A)\subset {\mathcal{C}}^{0}(A)$

[6.2.19] 

