(last updated: 3:33:16 PM, November 03, 2020)
Definition 4.1.1 (Local) Suppose that a function \(f:D\subset\mathbb R\rightarrow\mathbb R\). Then f is continuous at \(a\in D\) if and only if for any given \(\epsilon>0\), there exists \(\delta>0\) such that \(|f(x)-f(a)|<\epsilon\), provided that \(|x-a|<\delta\) and \(x\in D\).
Definition A point \(a\in Dom(f)\) is said to be an isolated point if it is not an accumulation point of \(Dom(f)\). The point \((a,f(a))\) of the function \(f\) (or a point on its graph) is said to be a singleton if \(a\) is not an accumulation point of \(Dom(f)\).
A function is always continuous at non-accumulation points, \(a\not\in acc(Dom(f))\) since you can always find a \(\delta>0\) small enough so that \((a-\delta,a+\delta)\) only contains \(a\) from the domain of \(f\). This indicates a departure from the naive notion that a function is continuous if you can draw its graph without lifting your pencil from the page!
Definition 4.1.2 (Global) A function \(f:D\subset\mathbb R\rightarrow\mathbb R\) is continuous on a set \(E\subset D\) if and only if \(f\) is continuous at each point (value of \(x\)) in \(E\). If \(f\) is continuous at every point in its domain, \(D\), we simply say that \(f\) is continuous.
Notice that the definition of continuity at \(x=a\) looks very similar to the definition of limit as \(x\) approaches \(a\) except that (1) the limiting value \(L\) is replaced by \(f(a)\), and (2) we now allow \(x= a\) (and do not require \(a\) to be an accumulation point of the domain).
Example: Consider \(f:\{0,1\}\rightarrow \mathbb R\) with \(f(0)=a\) and \(f(1)=b\neq a\). This function is only defined at these two points. Let \(\epsilon>0\) and \(\delta=1\). Then \(|x-0|<\delta=1\) and \(x\in Dom(f)\) implies that \(x=0\) necessarily. Thus \(|f(x)-f(0)|=|a-a|=0<\epsilon\). Hence \(f\) is continuous at \(x=0\). Similarly for \(x=1\), \(|x-1|<\delta=1\) and \(x\in Dom(f)\) implies that \(x=1\) necessarily and \(|f(x)-f(1)|=|b-b|=0<\epsilon\). So \(f\) is continuous at the only two points of its domain and is thus continuous by Definition 4.1.2.
The above example shows that the definition of continuity is different than the graph of the function being able to be drawn without lifting your pen from the page. In other words, a function can be continuous on a subset of \(\mathbb R\) but it’s graph not a contiguously drawn curve.
Example 4.1.4.
Every polynomial has domain \(\mathbb R\) and is continuous on its domain.
Definition 4.1.6 Suppose a function \(f : D\subset\mathbb R \rightarrow\mathbb R\). Then \(f\) is right continuous at \(a\), meaning that f is continuous from the right at \(a\) if and only if for any given \(\epsilon > 0\) there exists \(\delta > 0\) such that \(|f (x) - f (a) | <\epsilon\), provided that \(0 < x - a <\delta\) and \(x \in D\).
We can define left continuous similarly. I will also interchangeably say “continuous from the right at \(a\)” or make similar statements to indicate when a function is right continuous at a point.
A function can right continuous at \(a\) but not continuous as \(a\).
Example
Consider the Heaviside function: \[H(x)=\begin{cases}
0 &\text{ if } x<0\\
1 &\text{ if } x\geq0
\end{cases}
\]
This function is continuous from the right at \(x=0\) but has a discontinuity there.
Example
Consider the square root: \(f(x)=\sqrt x\)
This function is continuous from the right at \(x=0\) and is also continuous there.
Example
Consider the function: \(f(x)=\frac1{\sqrt x}\)
This function is not defined at \(x=0\) so there is no question of it being continuous there.
Note: see below as I make a departure from the textbook terminology standards here. This function is not defined at \(x=0\), so we cannot say whether it is continuous or discontinuous there. The function does nothing at \(x=0\) and therefor cannot be anything there. The function \(f\) does not exist at \(x=0\).
Definition (not in text, but consistent with text’s terminology usage) A function \(f\) is said to be discontinuous at \(a\) if
Although I depart from the textbooks terminology standards here, it is simply a matter of choice of language, a personal preference.
Boundedness:
Combining functions:
Composition of functions:
Theorem 4.1.9 Consider functions \(f : A\subset \mathbb R \rightarrow\mathbb R\) and \(g : B\subset \mathbb R \rightarrow\mathbb R\) such that \(f (A) \subset B\). If \(f\) is continuous at some \(x = a \in A\) and \(g\) is continuous at \(b = f (a) \in B\), then the function \(g \circ f\) is continuous at \(x = a\).
Example: Consider the function \[ f(x)=\begin{cases} 1 &\text{ if } x\in \mathbb Q\\ 0 &\text{ if } x\not\in \mathbb Q \end{cases} \] Consider \(\epsilon\leq 1\). Note that given any \(\delta>0\) the interval \((a-\delta,a+\delta)\) will have both rational and irrational numbers. So no matter which \(q\in Dom(f)\) we are considering, there will be an \(x\in(a-\delta,a+\delta)\) such that \(|f(x)-f(a)|=1\geq\epsilon\). Thus we cannot satisfy the definition of continuity at any \(a\in\mathbb R\).
This function is discontinuous at every real number!
Example: Consider the function \[ f(x)=\begin{cases} \frac1q &\text{ if } x=\frac pq\in \mathbb Q \text{ in reduced form, } p\in\mathbb Z,q\in\mathbb N\\ 0 &\text{ if } x\not\in \mathbb Q \end{cases} \] In my section 3.2 notes I show that \(\displaystyle\lim_{x\rightarrow a} f(x)=0\) for all \(a\in\mathbb R\). Since \(f(a)=0\) for irrational \(a\), we conclude that \(f\) is continuous at every irrational number. Since \(f(a)=1\) for all rational \(a\), we conclude that \(f\) is discontinuous at every rational number.
This is quite a strange function! This shows how strange structures can arise on the real line. I like to think of it as a consequence of packing infinitely many points into even the tiniest portions of the real line and the strange relationship between the rational and irrational numbers. These kind of mathematical contructs can have very counterintuitive properties.
\[ \diamond \S \diamond \]