# Real Analysis/Continuity

 Real Analysis Continuity

Now that we've defined the limit of a function, we're in a position to define what it means for a function to be continuous. The notion of Continuity captures the intuitive picture of a function "having no sudden jumps or oscillations". We will see several examples of discontinuous functions that illustrate the meaning of the definition. The idea of continuous functions is found in several areas of mathematics, apart from real analysis.

## Definition

Let $A\subseteq\mathbb{R}$
Let $f:A\to\mathbb{R}$
Let $c\in A$

We say that $f(x)$ is continuous at $c$ if and only if
For every $\varepsilon>0$, there exists $\delta>0$ such that $|x-c|<\delta\implies |f(x)-f(c)|<\varepsilon$.
We say $f$ itself is continuous if this condition holds for all points in $A$.

If $A$ is a union of intervals, the statement is equivalent to saying that $\lim_{x\to c}f(x)=f(c)$.

## Operations

Since limits are preserved under algebraic operations, let's check whether this is also the case with continuity.

### Algebraic

We see that if $f(x)$ and $g(x)$ are both continuous at c, continuity still works out fine for the following situations:

Addition $\lim_{x \rightarrow c}{(f+g)(x)} = f(c) + g(c)$ $\lim_{x \rightarrow c}{(f-g)(x)} = f(c) + g(c)$ $\lim_{x \rightarrow c}{(f \cdot g)(x)} = f(c) \cdot g(c)$ $\lim_{x \rightarrow c}{\lambda \cdot g(x)} = \lambda \cdot g(c)$ $\lim_{x \rightarrow c}{\left( \dfrac{1}{g} \right)(x)} = \dfrac{1}{g(c)}$ $\lim_{x \rightarrow c}{\left( \dfrac{f}{g} \right)(x)} = \dfrac{f(c)}{g(c)}$

Note that of course, for any division, g(c) must be a valid number i.e. not 0.

This is actually a corollary when you look at the proofs for the preservation of algebraic operation for limits. Simply replace the limit values L and M with ƒ(c) and g(c) respectively.

We can use sequential limits to prove that functions are discontinuous as follows:

• $f(x)$ is discontinuous at $c$ if and only if there are two sequences $(x_n)\rightarrow c$ and $(y_n)\rightarrow c$ such that $\lim_{n \rightarrow \infty}(f(x_n)) \not= \lim_{n \rightarrow \infty}(f(y_n))$.

### Composition

Composition is a lot trickier though, as always.

 Let $\epsilon > 0$ Since f is continuous, $\exists \delta_1 > 0: |x-c|<\delta_1 \implies |f(x)-f(c)|<\epsilon$. Since g is continuous, $\exists \delta_2 > 0: |x-c|<\delta_2 \implies |g(x)-g(c)|<\delta_1$. Thus $|x-c|<\delta_2 \implies |g(x)-g(c)|<\delta_1 \implies |f(g(x))-f(g(c))|<\epsilon$, so $(f \circ g)(x)$ is continuous on A. $\square$

But, it still works too, under continuity.

## The Three Continuity Theorems

Think about what an intuitive notion of continuity is. If you can’t the image of a polynomial function always works. The smooth curve as it travels through the domain of the function is a graphical representation of continuity. However, how do we mathematically know that it’s continuous? Well, we’ll start with the Three Continuity Theorems that will verify this notion.

### The Intermediate Value Theorem

This is the big theorem on continuity. Essentially it says that continuous functions have no sudden jumps or breaks.

Theorem
Let f(x) be a continuous function. If $a and $f(a), then $\exists c \in (a,b): f(c) = m$.
The intermediate value theorem: Given a continuous function on [a,b] and three variables a < c < b it must be the case that ƒ(a) < ƒ(c) < ƒ(b)

#### Proof

Let $S = \{x \in (a,b): f(x) < m\}$, and let $c = \sup S$.

Let $\epsilon = |f(c) - m|$. By continuity, $\exists \delta: |x-c|< \delta \implies |f(x)-f(c)|< \epsilon$.

If f(c) < m, then $|f(c+\frac{\delta}{2}) - f(c)| < \epsilon$, so $f(c+\frac{\delta}{2}) < f(c) + \epsilon = m$. But then $c + \frac{\delta}{2} \in S$, which implies that c is not an upper bound for S, a contradiction.

If f(c) > m, then since $c = \sup S$, $\exists x: x \in S, c>x>c-\delta$. But since $|x-c|<\delta$, $|f(x)-f(c)|<\epsilon$, so $f(x)> f(c) - \epsilon$ = m, which implies that $x \notin S$, a contradiction. $\Box$

We will now prove the Minimum-Maximum theorem, which is another significant result that is related to continuity. Essentially, it states that any continuous image of a closed interval is bounded, and also that it attains these bounds.

### Minimum-Maximum Theorem

This theorem functions as a first part in another bigger theorem. However, on its own, it helps bridge the gap between supremums and infimums in regards to functions.

Theorem
Given a continuous function ƒ on [a,b] i.e. $f:[a,b]\to\mathbb{R}$, if $a and $f(a), then $f([a,b])$ is bounded.

#### Proof

Assume if possible that $f$ is unbounded.

Let $x_1=\tfrac{a+b}{2}$. Then, $f$ is unbounded on at least one of the closed intervals $[a,x_1]$ and $[x_1,b]$ (for otherwise, $f$ would be bounded on $[a,b]$ contradicting the assumption). Call this interval $I_1$.

Similarly, partition $I_1$ into two closed intervals and let $I_2$ be the one on which $f$ is unbounded.

Thus we have a sequence of nested closed intervals $[a,b]\supseteq I_1\supseteq I_2\supseteq\ldots$ such that $f$ is unbounded on each of them.

We know that the intersection of a sequence of nested closed intervals is nonempty. Hence, let $x_0\in I_1\cap I_2\cap\ldots$

As $f(x)$ is continuous at $x=x_0$, there exists $\delta >0$ such that $x\in V_{\delta}(x_0)\implies f(x)\in (f(x_0)-1,f(x_0)+1)$ But by definition, there always exists $k\in\mathbb{N}$ such that $I_k\subseteq V_{\delta}(x_0)$, contradicting the assumption that $f$ is unbounded over $I_k$. Thus, $f$ is bounded over $[a,b]$

### Extreme Value Theorem

This is the second part of the theorem. It is the more assertive version of the previous theorem, stating that not only is there a supremum and a infimum, it also is reachable by the function ƒ and will be in between the interval you specified.

Theorem
Given a continuous function ƒ on [a,b] i.e. $f:[a,b]\to\mathbb{R}$, if $M,m$ are respectively the upper and lower bounds of $f([a,b])$, then there exist $c,d\in [a,b]$ such that $f(c)=M,f(d)=m$.
A depiction of the Extreme Value Theorem: Given a continuous function on [a,b], there must exist a maximum c and d such that ƒ(c) is the greatest value in the interval and ƒ(d) is the littlest.

#### Proof

Assume if possible, $M=\sup (f([a,b]))$ but $M\notin f([a,b])$.

Consider the function $g(x)=\frac{1}{M-f(x)}$. By algebraic properties of continuity, $g:[a,b]\to\mathbb{R}$ is continuous. However, $M$ being a cluster point of $f([a,b])$, $g(x)$ is unbounded over $[a,b$, contradicting (i). Hence, $M\in f([a,b])$. Similarly, we can show that $m\in f([a,b])$.

## Appendix

Continuity will come again in other branches in mathematics. You will come across not only different variations of continuity, but you will also come across different definitions of continuity too.

### Uniform Continuity

Let $A\subseteq\mathbb{R}$
Let $f:A\to\mathbb{R}$

We say that $f$ is Uniformly Continuous on $A$ if and only if for every $\varepsilon>0$ there exists $\delta>0$ such that if $x,y\in A$ and $|x-y|<\delta$ then $|f(x)-f(y)|<\varepsilon$

### Lipschitz continuity

Let $A\subseteq\mathbb{R}$
Let $f:A\to\mathbb{R}$

We say that $f$ is Lipschitz continuous on $A$ if and only if there exists a positive real constant $K$ such that, for all $x,y\in A$, $|f(x)-f(y)| \le K |x-y|$.

The smallest such $K$ is called the Lipschitz constant of the function $f$.

### Topological Continuity

As mentioned, the idea of continuous functions is used in several areas of mathematics, most notably in Topology. A different characterization of continuity is useful in such scenarios.

### Theorem

Let $A\subseteq\mathbb{R}$
Let $f:A\to\mathbb{R}$

$f(x)$ is continuous at $x=c$ if and only if for every open neighbourhood $V$ of $f(x)$, there exists an open neighbourhood $U$ of $x$ such that $U\subseteq f^{-1}(V)$

It must be mentioned here that the term "Open Set" can be defined in much more general settings than the set of reals or even metric spaces, and hence the utility of this characterization.