It is quite common for mathematicians to come up with more than one definition for a concept. Two definitions are said to be equivalent if a mathematical object satisfying either one of the definitions must also satisfy the other. The following are three equivalent definitions for continuity, one geometrical, one topological (based on open intervals), and one analytical (probably similar to one you saw in a calculus course).
You might review Definitions 1.25 and Definition 1.26 and the discussion following these definitions before reading the next definition.
The statement that the function \(f\) is continuous at the point \(p=(x,y)\) means that
\(p\) is a point on \(f\text{,}\) and
if \(H\) and \(K\) are any two horizontal lines with \(p\) between them, then there are two vertical lines, \(h\) and \(k\) with \(p\) between them so that if \(t\) is any point in the domain of \(f\) between \(h\) and \(k,\) then \((t,f(t))\) is in the rectangle bounded by \(h, k, H,\) and \(K\text{.}\)
The statement that the function \(f\) is continuous at the point \(p=(x,y)\) means that
\(p\) is a point on \(f\text{,}\) and
if \(S\) is any open interval containing the number \(f(x)\text{,}\) then there is an open interval \(T\) containing the number \(x\) such that if \(t \in T\text{,}\) and \(t\) is in the domain of \(f,\) then \(f(t) \in S\text{.}\)
The statement that the function \(f\) is continuous at the point \(p=(x,y)\) means that
\(p\) is a point on \(f\text{,}\) and
if \(\epsilon\) is any positive number, then there is a positive number \(\delta\) so that if \(t\) is in the domain of \(f\) and \(|t-x|\lt \delta,\) then \(|f(t) - f(x)| \lt \epsilon\text{.}\)
The statement that the function \(f\) is continuous at the number \(x\) means that \(x\) is in the domain of \(f\) and \(f\) is continuous at the point \((x,f(x))\text{.}\)
The statement that \(f\) is continuous function on the set \(M\) means that \(f\) is a function which is continuous at each number \(x\) in \(M\text{.}\)
Let \(f\) be the function such that \(f(x)=2\) for all numbers \(x>5\text{,}\) and \(f(x)=1\) for all numbers \(x \leq 5\text{.}\)
Show that \(f\) is not continuous at the point \((5,1)\text{.}\)
Show that if \(t\) is a number and \(t>5\text{,}\) then \(f\) is continuous at \((t,2)\text{.}\)
Show that if \(f\) is a function and \((x,f(x))\) is a point on \(f\text{,}\) and \(x\) is not a limit point of the domain of \(f\text{,}\) then \(f\) is continuous at \((x,f(x))\text{.}\)
Let \(f\) be the function such that \(f(x)=x^2\) for all numbers \(x\text{.}\) Show that \(f\) is continuous at the point \((2,4)\text{.}\)
If \(f\) is a function which is continuous on \([a,b]\) and \(x \in (a,b)\) such that \(f(x) > 0\) then there exists an open interval, \(T,\) containing \(x\) such that \(f(t) > 0\) for all \(t \in T\text{.}\)
If \(f\) is a function and \(p_1,p_2,p_3,\dots\) is a sequence of points in the domain of \(f\) converging to the number \(x\) in the domain of \(f\text{,}\) and \(f\) is continuous at \((x,f(x))\text{,}\) then \(f(p_1),f(p_2 ), \dots\) converges to \(f(x)\text{.}\)
The converse of this statement is that if \(f\) is a function so that for every sequence \(p_1, p_2, p_3, \dots\) in the domain of \(f\) converging to a point \(x\) we have that \(f(p_1),f(p_2 ), \dots\) converges to \(f(x)\) then \(f\) is continuous at \(x\text{.}\) This gives us a fourth equivalent definition for continuity of a function.
We say that a function \(f\) is continuous at the point \(x\) if and only if for every sequence \(p_1, p_2, p_3, \dots\) in the domain of \(f\) converging to \(x\) we have that \(f(p_1),f(p_2 ), \dots\) converges to \(f(x)\text{.}\)
Does there exists a function \(f\) that is defined on the open interval \((a,b)\text{,}\) continuous at a point \(p \in (a,b)\text{,}\) and a limit point of points at which \(f\) is not continuous?
Construct a function \(f\) that is defined on some closed interval \([a,b]\text{,}\) but not continuous at any point in \([a,b]\text{.}\)
If \(f\) and \(g\) are functions and there is a point common to the domain of \(f\) and the domain of \(g\text{,}\) then \(f+g\) denotes the function \(h\) such that for each number \(x\) in the domain of both of \(f\) and \(g\text{,}\) \(h(x)=f(x)+g(x)\text{.}\)
If each of \(f\) and \(g\) is a function, \(x\) is a point in the domain of each of \(f\) and \(g,\) \(f\) is continuous at the point \((x,f(x)),\) \(g\) is continuous at the point \((x,g(x)),\) and \(h=f+g\text{,}\) then \(h\) is continuous at the point \((x,h(x))\text{.}\)
We can't prove everything in the given time, but we'll assume additional theorems as needed about continuity. For example we'll assume that under appropriate conditions, the product, quotient, and composition of continuous functions are continuous and that all polynomials are continuous.
Suppose \(f\) and \(g\) are functions having domain \(M\) and each is continuous at the point \(p\) in \(M\text{.}\) Suppose that \(h\) is a function with domain \(M\) such that \(f(p)=h(p)=g(p)\) and for each number \(x\) in \(M\text{,}\) \(f(x) \le h(x) \le g(x)\text{.}\) Prove \(h\) is continuous at \(p\text{.}\)
(CA) If \(I_1,I_2,I_3, \dots\) is a sequence of closed intervals such that for each positive integer \(n\text{,}\) \(I_{n+1}\subseteq I_n,\) then there is a point \(p\) such that if \(n\) is any positive integer, then \(p\) is in \(I_n\text{.}\) In other words, there is a point \(p\) which is in all the closed intervals of the sequence \(I_1,I_2,I_3 , \dots\text{.}\)
(CA) If \(I_1,I_2,I_3 , \dots\) is a sequence of closed intervals so that for each positive integer \(n\text{,}\) \(I_{n+1} \subseteq I_n,\) and the length of \(I_n\) is less than \(1 \over n\text{,}\) then there is only one point \(p\) such that for each positive integer \(n\text{,}\) \(p \in I_n\text{.}\)
If \(p_1,p_2,p_3,\dots\) is a sequence of points in the closed interval \([a,b]\text{,}\) then there is a point in \([a,b]\) which is not in the sequence \(p_1,p_2,p_3,\dots\text{.}\)
If \(x\) is a limit point of the point set \(M,\) then there is a sequence of points \(p_1,p_2,p_3,\dots\) of \(M\text{,}\) all different and none equal to \(x\) which converge to \(x\text{.}\)
If \(x_1, x_2, x_3, \dots\) is a sequence of distinct points in the closed interval \([a,b],\) then the range of the sequence has a limit point.
A consequence of Theorem 2.21 is that every infinite bounded set has a limit point.
If \(f\) is a continuous function whose domain includes the closed interval \([a,b]\) and there is a point \(x\) in \([a,b]\) so that \(f(x)\) is greater than or equal to zero, then the set of all numbers \(x \in [a,b]\) such that \(f(x) \ge 0\) is a closed point set.
If \(f\) is a continuous function whose domain includes a closed interval \([a,b]\) and \(p \in [a,b]\text{,}\) then the set of all numbers \(x \in [a,b]\) such that \(f(x)=f(p)\) is a closed point set.
The statement that the point sets \(H\) and \(K\) are mutually exclusive or disjoint means that they have no point in common.
(CA) No closed interval is the union of two mutually exclusive closed point sets.
(CA) If \(f\) is a function with domain the closed interval \([a,b]\) and the range of \(f\) is \(\{-1,1\}\text{,}\) then there is a number \(x\) in \([a,b]\) at which \(f\) is not continuous.
(CA) Let \(f\) be a continuous function whose domain includes the closed interval \([a,b]\text{.}\) If \(f(a)\lt 0\) and \(f(b)>0\text{,}\) then there is a number \(x\) between \(a\) and \(b\) such that \(f(x)=0\text{.}\)
If \(f\) is a continuous function whose domain includes a closed interval \([a,b]\text{,}\) and \(L\) is a horizontal line, and \((a,f(a))\) is below \(L\text{,}\) and \((b,f(b))\) is above \(L\text{,}\) then there is a number \(x\) between \(a\) and \(b\) such that \((x,f(x))\) is on \(L\text{.}\)