Karl's Calculus Tutor - Box 3.0e: Solution to Continuity Proof Problem

Box 3.0d: Solution to Continuity Proof Problem

The Stepping Stone Theorem

The Stepping Stone Theorem says that a function, f(x), is continuous at a point, x = c, if and only if for every sequence of real numbers, x₁, x₂, x₃, ... , whose limit is c, the limit of the sequence, f(x₁), f(x₂), f(x₃), ... , is exactly f(c). In other words, if the stepping stones lead to the riverbank, then the moss on the stepping stones leads to the moss on the riverbank. You only needed to prove the "only if" part of this, but I shall give you the proofs to both parts here.

Outline of Proof to the "only if" Part

We show that if the sequence converges to a limit of c, then as the subscripts grow, the x_n's must fall within smaller and smaller intervals of c. Another way of saying that is that they will eventually fall within d of c, no matter how close to zero you choose d to be. There is not much to show here since this is all part of how we define what it means for a sequence to converge to a limit. We then go on to show that applying the definition of continuity to this leads to the convergence of f(x₁), f(x₂), f(x₃), ... to a limit of f(c).

Proof of the "only if" Part

If the sequence, x₁, x₂, x₃, ... , converges to c, then if I tell you that x_n has got to be within d of c, then you can tell me how big n has to be to guarantee that. That is the definition of convergence of a sequence.

We also have...

If f(x) is continuous at c, then for any e > 0, no matter how small, there is a d that makes

   |f(x) - f(c)|  £  e

whenever |x - c| £ d. That is the definition of continuity in its delta-epsilon form.

Whatever e I choose, you can find a d small enough to make the above inequality true. So every x within d of c gives you an f(x) that is within e of f(c). But whatever that d is, you can also find an n big enough so that x_k is within d of c whenever k ³ n. And that means that f(x_k) must be within e of f(c) whenever k ³ n. This is exactly how we define convergence of a sequence to a limit. And that is true no matter how small an e I chose in the first place. So it follows that the sequence, f(x₁), f(x₂), f(x₃), ... , must converge to a limit of f(c).

Outline of Proof of the "if" Part (optional material)

Here we must show that if every set of stepping stones that leads to the river bank has moss that leads to the moss on the riverbank, then the moss function is continuous. Or in other words, if the moss function is not continuous, then at least one set of stepping stones that leads to the riverbank has moss that does not lead to the moss on the riverbank. That is the approach this proof takes. We assume that the function, f(x), is discontinous at x = c. From that assumption we find a sequence, x₁, x₂, x₃, ... , that converges to a limit of c, but when you take the sequence, f(x₁), f(x₂), f(x₃), ... , it fails to converge to a limit of f(c). Since the premise was that every such sequence should converge to f(c), finding one that doesn't proves that the assumption that f(x) was discontinuous at c had to be false.

Proof of the "if" Part (optional material)

First we define a sequence of d's. Let

          1
   d _n  =   
          n

Assume now that f(x) is discontinuous at x = c. That means that there exists some e > 0 small enough so that for every d > 0, no matter how small, there is a point, x, such that both

   |x - c|  £  d

and

   |f(x) - f(c)|  >  e

In other words, for f(x) to be discontinous at c, no matter how close I require an x to be to c, you can always find such an x that will make f(x) farther than e from f(c).

Since that rule exists for every d, that means you can find such an x that is contained in the interval, c - d₁ £ x £ c + d₁. Call it x₁. You can also find one that is contained in the interval, c - d₂ £ x £ c + d₂. Call it x₂. And in general, you can find such an x contained in the interval, c - d_n £ x £ c + d_n, no matter how big n is. Call it x_n.

There is no question that the d_n's converge to zero. That means that as n gets big, x_n gets squeezed closer and closer to c. In fact the x_n's must converge to c. But remember that each x_n was chosen so that f(x_n) would be farther than e from f(c). The discontinuity at c was what allowed you to choose it that way. And because f(x_n) never gets within e of f(c), we know that the sequence, f(x₁), f(x₂), f(x₃), ... , cannot converge to a limit of f(c). That contradicts the premise of the theorem, and so, f(x) must be continuous at x = c.

Comments

I don't expect that your instructor will ask you to reproduce the "if" part on an exam, but he or she may very well ask you to reproduce the "only if" part -- that is showing that f(x) being continuous at c implies that every sequence, x_n, that converges to c generates a sequence, f(x_n), that converges to f(c). Be sure you know what the theorem means, because can be worded in a number of different ways. And learn the proof.

I included the proof of the "if" part only to satisfy the curiosity of those who are interested.

One more optional item for the curiosity of those who are interested. We defined the property of continuity using a delta-epsilon contract. We have now shown that this converging sequence property given in the stepping stone theorem is true if and only if the continuity property is true. When you have one property that is true if and only if another is true, the two properties are said to be equivalent. That means that we could have chosen to define continuity using the converging sequence property, gone on to prove the delta-epsilon property, and everything else about continuity would remain the same.

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