Continuity

1. Continuity

So far, we have proved three main results concerning limits and arithmetical operations + , · , - and (with one important proviso) a similar result for divison. Because of the algebraic nature of addition, multiplication and division, such results are often grouped together under the title The Algebra Of Limits. I find this name very unfortunate for a number of reasons: first, it suggests the work here has been algebraic (it hasn't: it has been analytic); second, it suggests that the results are true because of the special algebraic nature of the operations on limits being considered; third, the initials are "AOL".

To really understand what is going on here we need to take a slightly more abstract view and look to see what addition, subtraction and multiplication have in common. These are all functions, defined on the real numbers, with two inputs or arguments. If f ( x , y ) is the function x + y , or x - y , or x · y then f : 2 is a function with the following property. For all sequences ( a n ) and ( b n ) we have,

lim n f ( a n , b n ) = f ( lim n a n , lim n b n )

Where, by this we mean: if the limits on the right hand side exist in the reals then the limit on the left exists and equals the expression given. This is the main content of the last two web pages on sums of sequences and products of sequences.

Other functions have this property such as the distance function d ( x , y ) = | x - y | . In fact there is no need for the function to have exactly two arguments, and the idea works just as well with one, two, three or more arguments. Some functions do not have the property though. Consider the integer part function, [ x ] . If we define a sequence ( a n ) by a n = 1 - 1 n then lim n a n = 1 by the convergence of the constant sequence 1 to 1 , the convergence of 1 n to 0 , and the result mentioned above on subtraction, that lim n b n - c n = lim n b n - lim n c n where b n = 1 and c n = 1 / n . But [ a n ] = [ 1 - 1 n ] = 0 for all n , so lim n [ a n ] = 0 [ lim n a n ] . In other words, limits cannot be pushed through integer-part signs. Examining the graph of y = [ x ] gives some intuition as to the reason. This graph has a jump at x = 1 . In other words it is not continuous.

Definition.

Let f : be a function and l . We say that f is continuous at l if whenever ( a n ) is a sequence that converges to l then the sequence ( b n ) defined by b n = f ( a n ) also converges and its limit is equal to f ( l ) . We say that f is continuous if it is continuous at all l .

So for example, the function f ( x ) = [ x ] is not continuous at l for l = , -2 , -1 , 0 , 1 , 2 , but the functions g ( x ) = 2 x and f ( x ) = x 2 are continuous at all values of l since lim n 2 a n = 2 ( lim n a n ) and lim n a n 2 = ( lim n a n ) 2 for all convergent sequences ( a n ) .

Definition.

Let f : 2 be a function with two arguments and l , m . We say that f is continuous at ( l , m ) if whenever ( a n ) and ( b n ) are sequences that converge to l , m respectively then the sequence ( c n ) defined by c n = f ( a n , b n ) also converges and its limit is equal to f ( l , m ) . We say that f is continuous if it is continuous at all ( l , m ) 2 .

Continuity is about pushing functions through limits and the most important continuous functions are the arithmetic ones such as plus and times. The following theorem covers the most important cases of these:

Continuity of plus, times, etc.

  1. The addition function + with two real arguments is continuous.
  2. The subtraction function - with two real arguments is continuous.
  3. The multiplication function · with two real arguments is continuous.
  4. The division function / (i.e., the function defined by f ( x , y ) = x / y ) with two real arguments is continuous at all ( l , m ) 2 except when m = 0 .
  5. The square root function is continuous at all positive l .
  6. The absolute value function | | is continuous.

The first four parts of this have already been proved. We sketch the proof of the others here.

Proposition.

The square root function x is continuous at all l > 0 .

Proof.

Let a n l as n , where the limit l is positive. The by a previous proposition there is N 0 such that a n > | l | / 2 for all n N 0 . Thus the sequence ( a n ) is meaningful. We show that this sequence converges to l .

Subproof.

Let ε > 0 be arbitrary.

Subproof.

Let N 1 such that n N 1   | a n - l | < ε l .

Let N = max ( N 0 , N 1 ) .

Subproof.

Let n N be arbitrary.

Then a n > 0 and | a n - l | = | a n - l | a n + l < ε l l = ε .

So n N   | a n - l | < ε .

So N   n N   | a n - l | < ε .

So ε > 0   N   n N   | a n - l | < ε .

Proposition.

The absolute value function | | is continuous at all l .

Proof.

We have to prove that a n l implies | a n | | l | . We do this in the case when l > 0 . The other cases ( l < 0 and l = 0 ) are easy variations left as exercises.

Subproof.

Let ε > 0 be arbitrary.

Subproof.

Let N 0 such that n N 0   a n > l / 2 .

Let N 1 such that n N   | a n - l | < ε .

Let N = max ( N 0 , N 1 ) .

Subproof.

Let n N be arbitrary.

Then a n and l are both positive and so | a n | = a n , | l | = l , so | | a n | - | l | | = | a n - l | < ε .

So n N   | | a n | - | l | | < ε .

So N   n N   | | a n | - | l | | < ε .

So ε > 0   N   n N   | | a n | - | l | | < ε .

These results can be combined to give other continuous functions.

The distance function d ( x , y ) = | x - y | is continuous at all ( l , m ) 2 .

Proof.

Let a n l and b n m . Then d ( a n , b n ) = | a n - b n | . So, taking limits as n ,

lim d ( a n , b n ) = lim | a n - b n | = | lim a n - b n | ,

as | | is continuous, and this limit equals

| lim a n - lim b n | = d ( lim a n , lim b n ) ,

as - is continuous.

2. Examples

Continuity of the familiar functions we see here is useful in computing limits of what would otherwise be rather scary-looking sequences.

Example.

Let ( a n ) be defined by

a n = 3 n 3 + 2 + 2 n ( n + 1 ) 4 - n 4

We compute its limit as n as follows.

a n = 3 n 3 + 2 + 2 n 4 n 3 + 6 n 2 + 4 n + 1 = 3 + 2 n -3 + 2 n -1 / 2 4 + 6 n -1 + 4 n -2 + n -3 3 + 0 + 0 4 + 0 + 0 + 0 = 3 2

as n -1 / 2 , n -1 , n -2 , n -3 all converge to 0 , as + and scalar multiplication are continuous everywhere, as is continuous at 3 and 4 and division is continuous at ( 3 , 4 ) .

Example.

Let the sequence ( a n ) be defined by a 1 = a 2 = 1 , and

a n + 2 = 3 a n + 2 a n + 1 a n + 5 a n + 1

Let us suppose that a n l 0 as n . Then ( a n + 1 ) and ( a n + 2 ) are subsequences of ( a n ) so by our assumption should tend to the same limit l . By continuity of addition, scalar multiplication everywhere and division at ( 5 l , 6 l ) (noting our assumption l 0 ) we have

3 a n + 2 a n + 1 a n + 5 a n + 1 3 l + 2 l l + 5 l = 5 6

But a n + 2 l so by the uniqueness of limits we must have l = 5 6 .

The conclusion is that if the sequence ( a n ) converges to some l 0 then l = 5 6 . Of course this assumption is a big one and sequences defined in this sort of way need not converge at all. We therefore need to find ways to prove that a sequence converges without necessarily finding the limit: this is the subject of the next block of work.

3. Summary

Pushing limits through functions is a natural idea and can save an enormous amount of time in calculations and proofs. You have seen in this web page how it can be justified for some functions and not for others. The functions for which we can manipulate limits in this way are called continuous. Sometimes a function is continuous at some points but not others. The integer-part function and the division function are examples of these.

Implicitly assuming a function is continuous and pushing a limit through is one of the most common sources of errors in a mathematical argument. There are many fallacious arguments that prove that 0 = 1 or some such absurdity that reply on assuming a non-continuous function is continous to trick the reader. Please always state in your own work reasons for such arguments (and check the functions really are continuous).