6.3. Calculation Rules for Continuous Functions

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

If g is continuous at a and if  f is continuous at ${\displaystyle g(a)\in A}$ then

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

• ${\displaystyle A\ni a_n\to a\Rightarrow f(a_n)\to f(a)}$
• ${\displaystyle B\ni a_n\to a\Rightarrow g(a_n)\to g(a)}$

Thus the first limit theorem assures that

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

Proposition (Global Continuity Theorems):  

Proposition:  For any bijection  ${\displaystyle f:I\to B\subset ℝ}$ 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 ${\displaystyle \infty }$ for the right and ${\displaystyle -\infty }$ for the left boundery. Thus ${\displaystyle ℝ}$ and e.g. ${\displaystyle {ℝ}^{>0}}$ are regarded as intervals as well.

we have:

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

to get a subsequence ${\displaystyle (f^{-1}(a_{k_n}))}$ in I with all it's members being outside ${\displaystyle ]f^{-1}(a)-\epsilon ,f^{-1}(a)+\epsilon [}$ [1]. As I is closed ${\displaystyle (f^{-1}(a_{k_n}))}$ 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 ${\displaystyle (f^{-1}(a_{k_{j_n}}))}$ such that

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

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