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We recall that *uniform convergence* is convergence in the metric *d*_{∞} on a space of functions.

Some of the examples we considered in the last section show that it is possible for a sequence of continuous functions to be pointwise convergent to a non-continuous function.

Some of the other metrics we have considered give some similar unpleasant results.

For example, consider the sequence of functions (*f*_{n}) where *f*_{n} is:

Then the pointwise limit of (*f*_{n}) is the function *f* given by *f* (*x*) = 0 if *x* < ^{1}/_{2} and *f* (*x*) = 1 otherwise. Its graph is:

Then it is easy to see that *d*_{1}(*f*_{n}, *f*) = ^{1}/_{2n}→ 0 as *n*→ ∞. Unfortunately, *f* does not lie in *C*[0, 1] and so we cannot quite talk about convergence in this space. However, the sequence (*f*_{n}) is a *Cauchy sequence* in *C*[0, 1] with the metric *d*_{1}.

This motivates:

**Definition**

**Remarks**

- What we saw earlier about the Completeness axiom can now be phrased as:

*The real numbers***R***form a complete space under the usual metric.* - The last example shows that
*C*[0, 1] under the metric*d*_{1}is*not*a complete space. We saw earlier that the set of rationals**Q**under the usual metric is not a complete space.

The nice thing about the metric *d*_{∞} is:

**Theorem**

- We need to show that if (
*f*_{n}) is a Cauchy sequence under*d*_{∞}(that is: Given*ε*> 0 there exists*N*such that if*m*,*n*>*N*then*d*_{∞}<*ε*) then the sequence has a limit in*C*[0, 1].First we show that the sequence converges to its pointwise limit in the metric

*d*_{∞}.**Proof of that**

Take a point*x*∈ [0, 1]. Then the sequence (*f*_{n}(*x*)) is a real sequence which, from the fact that (*f*_{n}) is a Cauchy sequence, is a Cauchy sequence in**R**and hence has a limit which we will call*f*(*x*). Thus the pointwise limit*f*is defined.Since at each point

*f*_{n}(*x*) is close to*f*(*x*) we have the lub{ |*f*_{n}(*x*) -*f*(*x*)| for*x*∈ [0, 1] } is small and so (*f*_{n})→*f*in*d*_{∞}.The last thing we need to prove is that the pointwise limit

*f*is continuous.**Proof of that**

To prove that*f*is continuous at*p*∈ [0, 1], take some*ε*> 0. Choose*N*such that*d*_{∞}(*f*_{n},*f*) <*ε*.

Then since*f*_{n}is continuous, there exists*δ*such that if*x*is in the interval (*p*-*δ*,*p*+*δ*) then |*f*_{n}(*x*) -*f*_{n}(*p*)| <*ε*. Then |*f*(*x*) -*f*(*p*)| = |*f*(*x*) -*f*_{n}(*x*) +*f*_{n}(*x*) -*f*_{n}(*p*) +*f*_{n}(*p*) -*f*(*p*)| < |*f*(*x*) -*f*_{n}(*x*)| + |*f*_{n}(*x*) -*f*_{n}(*p*)| + |*f*_{n}(*p*) -*f*(*p*)| <*ε*+*ε*+*ε*and so can be made arbitrarily small.

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