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Continuity Support

Continuity

Definition

Let $S \subseteq \R,$ $c \in S,$ and let $f \colon S \to \R$ be a function. We say that $f$ is continuous at $c$ if for every $\epsilon \gt 0$ there is a $\delta \gt 0$ such that whenever $x \in S$ and $\abs{x-c} \lt \delta,$ then $\abs{\,f(x)-f(c)} \lt \epsilon.$

When $f \colon S \to \R$ is continuous at all $c \in S,$ then we simply say $f$ is a continuous function.

Explore the definition of continuity: Drag the slider and then move the top point on $y$-axis.

Consider a function $f \colon S \to \R$ defined on a set $S \subseteq \R$ and let $c \in S.$ Then

  1. If $c$ is not a cluster point of $S,$ then $f$ is continuous at $c.$
  2. If $c$ is a cluster point of $S,$ then $f$ is continuous at $c$ if and only if the limit of $f(x)$ as $x \to c$ exists and \begin{equation*} \lim_{x\to c} f(x) = f(c) . \end{equation*}
  3. The function $f$ is continuous at $c$ if and only if for every sequence $\{ x_n \}$ where $x_n \in S$ and $\lim\, x_n = c,$ the sequence $\{ \,f(x_n) \}$ converges to $f(c).$
Let's prove item 1. Suppose $c$ is not a cluster point of $S.$ Then there exists a $\delta > 0$ such that \[S \cap (c-\delta,c+\delta)= \{ c \}.\]

For any $\epsilon > 0,$ simply pick this given $\delta.$ The only $x \in S$ such that $\abs{x-c} \lt \delta$ is $x=c.$ \[\Ra \;\abs{\,f(x)-f(c)} = \abs{\,f(c)-f(c)} = 0 \lt \epsilon.\]

Now let's prove item 2. Suppose $c$ is a cluster point of $S.$ $\nec$ First suppose $f$ is continuous at $c.$ For every $\epsilon > 0,$ there exists a $\delta \gt 0$ such that for $x \in S$ where $\abs{x-c} \lt \delta\;$ $\Ra \;\abs{\,f(x)-f(c)} \lt \epsilon.$ This is true since $x \in S \setminus \{ c \} \subset S.$ Therefore, $\lim_{x\to c} f(x) = f(c).$

Now let's prove the other direction. Again we assume $c$ is a cluster point of $S.$ $\suf$ Now assume that the limit of $f(x)$ as $x \to c$ exists and $\lim_{x\to c} f(x) = f(c).$ Then for every $\epsilon > 0,$ there is a $\delta \gt 0$ such that if $x \in S \setminus \{ c \}$ and $\abs{x-c} \lt \delta\,$ $\,\Ra \abs{\,f(x)-f(c)} \lt \epsilon.$ In particular $\abs{\,f(c)-f(c)} = 0 \lt \epsilon,$ and the definition of continuity at $c$ is satisfied.


Discontinuity criterion

Let $S\subseteq \R,$ let $f \colon S \to \R$ and let $c\in S.$ Then $f$ is discontinuous at $c$ if and only if there exists a sequence $\{x_n\}$ in $S$ such that $\lim x_n = c,$ but the sequence $\left\{\,f(x_n)\right\}$ does not converge to $f(c)$.
$\displaystyle f(x)=\sin\frac{1}{x},$ for $x\neq 0$ and $f(x)=0$ for $x=0.$
We can find a sequence $x_n$ such that $x_n\ra 0$ as $n\ra \infty.$
C
Notice how the sequence $f\left(x_n\right)$ does not converge to $0.$

Basic properties

Let $f \colon S \to \R$ and $g \colon S \to \R$ be functions continuous at $c \in S.$
  1. The function $h \colon S \to \R$ defined by $h(x) := f(x)+g(x)$ is continuous at $c.$
  2. The function $h \colon S \to \R$ defined by $h(x) := f(x)-g(x)$ is continuous at $c.$
  3. The function $h \colon S \to \R$ defined by $h(x) := f(x)g(x)$ is continuous at $c.$
  4. If $g(x)\not=0$ for all $x \in S,$ the function $h \colon S \to \R$ defined by $h(x) := \dfrac{f(x)}{g(x)}$ is continuous at $c.$

Claim: The functions $\sin(x)$ and $\cos(x)$ are continuous.

Proof.
Recall that $\abs{\sin(x)} \leq \abs{x},$ $\abs{\cos(x)} \leq 1,$ and $\abs{\sin(x)} \leq 1.$ Then

\begin{eqnarray*} \abs{\sin(x)-\sin(c)} & = & \abs{2 \sin \left( \frac{x-c}{2} \right) \cos \left( \frac{x+c}{2} \right)} \\ & = & 2 \abs{ \sin \left( \frac{x-c}{2} \right) } \abs{ \cos \left( \frac{x+c}{2} \right) }\\ & \leq & 2 \abs{ \sin \left( \frac{x-c}{2} \right) }\\ & \leq & 2 \abs{ \frac{x-c}{2} } = \abs{x-c}. \end{eqnarray*}

On the other hand we have

\begin{eqnarray*} \abs{\cos(x)-\cos(c)} & = &\abs{-2 \sin \left( \frac{x-c}{2} \right) \sin \left( \frac{x+c}{2} \right)} \\ & = & 2 \abs{ \sin \left( \frac{x-c}{2} \right) } \abs{ \sin \left( \frac{x+c}{2} \right) }\\ & \leq & 2 \abs{ \sin \left( \frac{x-c}{2} \right) }\\ & \leq & 2 \abs{ \frac{x-c}{2} } = \abs{x-c}. \end{eqnarray*}

Using previous inequalities and applying the definition of continuity we can conclude that the functions $\sin(x)$ and $\cos(x)$ are continuous.


Uniform continuity

Let $S \subseteq \R,$ and let $f \colon S \to \R$ be a function. Suppose for every $\epsilon \gt 0$ there exists a $\delta > 0$ such that whenever $x, c \in S$ and $\abs{x-c} \lt \delta,$ then $\abs{\,f(x)-f(c)} \lt \epsilon.$ Then we say $f$ is uniformly continuous.
Geometric interpretation: Uniformly continuos vs Nonuniformly continuous.
$f \colon [0,1] \to \R$ defined by $f(x) := x^2$ is uniformly continuous.

Proof. Note that $0 \leq x,c \leq 1.$ Then

\begin{eqnarray*} \sabs{x^2-c^2} & = &\sabs{x+c}\sabs{x-c} \\ & \leq & \bigl(\sabs{x}+\sabs{c}\bigr) \sabs{x-c} \\ & \leq & (1+1)\sabs{x-c} . \end{eqnarray*}

Therefore, given $\epsilon > 0,$ let $\delta := \dfrac{\epsilon}{2}.$ If $\sabs{x-c} \lt \delta,$ then $\sabs{x^2-c^2} \lt \epsilon.$

On the other hand, $g \colon \R \to \R$ defined by $g(x):=x^2$ is not uniformly continuous.

Proof. Suppose it is uniformly continuous, then for every $\epsilon > 0,$ there would exist a $\delta > 0$ such that if $\sabs{x-c} \lt \delta,$ then $\sabs{x^2 -c^2} \lt \epsilon.$ So consider $\epsilon =1.$ If such $\delta$ existed and $c = x+ \delta/2$ $\;\Ra \ds\sabs{x^2 - \left(x+ \frac{\delta}{2}\right)^2} \lt 1.$ However $ \ds \abs{x\delta + \dfrac{\delta^2}{4}} \lt 1$❗ which is a contradiction, since we can choose $x$ large.

The function $f \colon (0,1) \to \R$ defined by $f(x) := \dfrac{1}{x}$ is not uniformly continuous.

Proof. Choose $\epsilon=2$ and $0 \lt \delta.$ Set $\delta_0 = \min\{\delta/2, 1/4\}$, $x= \delta_0$, and $y = 2 \delta_0.$ Then $x,y\in(0,1)$ and $\abs{x-y}= \delta_0 \lt \delta$ but

\begin{eqnarray*} \abs{\frac{1}{x}-\frac{1}{y}} = &\abs{\frac{y-x}{xy}} = \abs{\frac{\delta_0}{2\delta_0^2} } = \ds\abs{ \frac{1}{2\delta_0}} \geq 2 = \epsilon. \end{eqnarray*}

Hence $f$ is not uniformly continuous.

Geometric interpretation: $f(x) = \dfrac{1}{x}$ is not uniformly continuous.



More coming soon!