# 16.1 Limits and an Introduction to Point Set Topology

Mathematicians have, especially since the 19th Century, wanted to make the subject of calculus rigorous, which means completely logically defined. We have called a function differentiable at a point if its graph "looks like a straight line" at that point. This is perhaps intuitive, but is certainly more intuitive than rigorous.

To introduce rigor we define the notion of the limit of a sequence.

An infinite sequence of numbers is said to converge, if for any criterion, (say a zillionth, whatever that means), beyond some point in the sequence, the difference between any two entries is less than that criterion.

The sequence $$\{\frac{1}{n}\}$$ converges because beyond the zillionth term the difference between any two entries is less than a zillionth.

A sequence of numbers in a given set converges to a limit $$x$$ if it converges and the difference between any entry and $$x$$ is less than that criterion, beyond some point.

The sequence $$\{\frac{1}{n}\}$$ converges to $$0$$.

We say that a function $$f$$ is continuous at argument $$A$$ if its values on any sequence of numbers that converge to $$A$$, converge to its value at $$A$$. We write this as $$\lim_{x \to A} f(x) = f(A)$$.

We say that $$f$$ is differentiable at $$A$$ if for any sequence of numbers $$\{d_j\}$$ none of them $$0$$ that converge to $$0$$, the limit as $$d$$ approaches $$0$$ of $$\frac{f(A+d_j)-f(A)}{d_j}$$ ie $$\lim_{d_j \to 0} \frac{f(A+d_j)-f(A)}{d_j}$$ exists, and we call that limit the derivative of $$f$$ at $$A$$, which we have written as both $$f'(A)$$ and $$\frac{df(A)}{dA}$$.

There are a lot of wonderful concepts and definitions that can be made once limits are defined.

A set $$S$$ of numbers is said to be complete, if every convergent sequence of elements of $$S$$ converges to a number in $$S$$.

The limit points of a set $$S$$ are those numbers that are limits of sequences of members of that set.

A set is closed if it contains all its limit points.

Notice that $$0$$, by definition is not a positive number, so that there are sequences of positive numbers that do not converge to a positive number, because they converge to $$0$$. Thus the positive numbers are not closed.

Remember that rational numbers are those which, beyond some point endlessly repeat some sequence of decimal digits.

Consider a number which, after the decimal point, starts with $$1$$ and has a sequence of zeroes and $$1$$'s that has exactly $$j$$ consecutive zeroes after the $$j^{th}$$ occurance of $$1$$. This number does not repeat after some point, and so it is not rational. But it is the limit of the sequence gotten by replacing all its entries beyond the $$n^{th}$$ by zeros (as $$n$$ goes to $$+\infty$$), all of which are rational, each ending in repeating $$0$$'s.

Thus, the rational numbers are not closed.

A boundary point of a set $$S$$ of real numbers is one that is a limit point both of $$S$$ and the set of real numbers not in $$S$$. Thus, if $$S$$ is the interval of points between $$a$$ and $$b$$ including the endpoints $$a$$ and $$b$$, then $$a$$ and $$b$$ are its boundary points. This $$S$$ is closed, because it contains all possible of its limit points.

An open set is one that contains no boundary points. The interval of points between $$a$$ and $$b$$ not including its endpoints is open. If an interval is defined to contain only one of its endpoints, it is neither open nor closed.

Closed sets are complements of open sets. Since closed sets contain all their boundary points, their complements, which contain all points of the space not in them, contain none of them.

What infinite sequences do not converge?

Non-convergence always occurs when a sequence of real numbers is unbounded: for example, the sequence

$$1,2,\dots n, \dots$$, does not converge.

Also, bounded sequences that have more than one limit point do not converge. For example

$$1,0,1,0,1,0,\dots$$ has limit points $$1$$ and $$0$$, and does not converge as the difference between consecutive terms is always $$1$$, and never gets below $$1$$.

A set in which every sequence of its elements has at least one limit point inside it is said to be sequentially compact. To be sequentially compact a set $$S$$ must be closed, or else, by definition, there is a convergent sequence of its elements that does not converge to a member of $$S$$. $$S$$ must be bounded or else there is a sequence that grows indefinitely with no finite limit point. (For example, choose as $$(n+1)^{st}$$ member of the sequence the smallest of its elements that is at least one greater than its $$n^{th}$$ member.)

On the other hand, if a set $$S$$ of real numbers is closed and bounded, every sequence of elements of $$S$$ has at least one limit point in $$S$$.

This statement follows from two observations, which we will prove. First, if $$S$$ is bounded, a sequence of its elements whose members increase, so that the $$(n+1)^{st}$$ member is at least as big as the $$n^{th}$$, must converge.

If $$S$$ is closed, an increasing sequence must converge to the least upper bound of its members, which will be an element of $$S$$ and that will be a limit point. And the same is true for a decreasing sequence.

Second, every infinite sequence must contain either an infinite subsequence that is increasing, or one that is decreasing or both.

Together, these statements mean that any infinite sequence of real numbers that are bounded and lie in a closed set, have a convergent sequence and hence a limit point.

We prove the first: The least upper bound of a set $$Q$$ of numbers is the smallest number that is at least as big as every element of $$Q$$. If $$Q$$ consists of the members of an increasing sequence, this least upper bound must be a limit point of the sequence. It certainly cannot be less than any member of the sequence. And if it is a zillionth greater than all members, it is not their least upper bound. This proves the first observation.

There is an elegant proof of the second observation, gotten by considering a finite sequence of numbers, say of length $$N$$. We show that every such sequence has either an increasing or decreasing subsequence of length at least the square root of $$N$$. Since the square root of an infinite number is still infinite, this result tells us that any infinite sequence must have an infinite increasing or decreasing subsequence either of which must have a limit point.

To show this, starting at the beginning of the sequence, keep track of the length of the longest increasing and longest decreasing sequence which ends at each member. The first such pair will be $$(1,1)$$, then $$(2,1)$$ or $$(1,2)$$ depending on whether the second member is bigger or smaller that the first (if it is the same as the first we would get $$(2,2)$$).

The wonderful fact is that no two members can have the same pair of numbers. If say, some member, $$m$$ has pair $$(i,j)$$ then any subsequent member, $$q$$, will get at least pair $$(i+1,j)$$ or $$(I,j+1)$$ because there will be an increasing sequence gotten by adding $$q$$ to the old increasing sequence ending at $$m$$ if $$q$$ is bigger than $$m$$, and a decreasing sequence gotten by adding $$q$$ similarly if $$q$$ is less than $$m$$.

Our claim follows from the fact that the number of distinct ordered pairs of positive integers both less than $$n$$ is $$(n-1)^2$$. Another way to express this fact is that there must be a "monotone" sequence of length at least $$n$$ among the first $$(n-1)^2 + 1$$ members of any sequence.

The two claims together tell us that any bounded closed set of real numbers is sequentially compact.

Exercise: Similar results hold for sets of ordered pairs, (or ordered $$n$$ tuples) of real numbers, which correspond to points in a two dimensional space (or $$n$$ dimensional space). Generalize the definitions above to apply to such sets, and prove that any bounded closed set of such pairs is sequentially compact.