Let’s consider another question: suppose *f* : *D* → **R** is continuous, where *D* is a subset of **R**. If (*x _{n}*) is a sequence in

*D*converging to some real

*L*, is it true that (

*f*(

*x*)) is also convergent? Now if

_{n}*L*is in

*D*, then we know that (

*f*(

*x*)) → (

_{n}*f*(

*L*)). But what if it’s not?

**Examples**

- Let
*f*(*x*) = 1/*x*, defined on (0, ∞). Take the sequence*x*= 1/_{n }*n*which converges to 0. Then the resulting sequence (*f*(*x*)) = (_{n}*n*) diverges. - Let
*f*(*x*) = sin(1/*x*), defined on (0, ∞). Then the sequence*x*= 2/((2_{n }*n*-1)π) converges to 0 but the resulting (*f*(*x*)) = (1, -1, 1, -1, … ) diverges._{n}

## Closed Subsets

Let’s first consider the case where the limit *L* must lie in *D*. To that end, we have the following result.

Proposition. For an arbitrary subset , the following are equivalent.

- If (x
_{n}) is a sequence in D converging to L, then L is in D.- D contains all its points of accumulations.
- The complement D
^{c}=R-D is an open subset ofR.

**Proof**.

(1)→(2) : Let *x* be a point of accumulation of *D*. Hence for each ε = 1, 1/2, 1/3, …, there exists , such that 0 < |*x _{n}*–

*x*| < 1/

*n*for each

*n*. Then (

*x*)→

_{n}*x*must be in

*L*.

(2)→(3) : Let . Since *a* is not in *D* it’s not a point of accumulation of *D*. By definition, this means that there exists ε>0 such that the set of *x* satisfying 0 < |*x*–*a*| < ε does not intersect *D*. Together with the fact that *a* is not in *D*, we see that , so *D ^{c}* is open.

(3)→(1) : Let (*x _{n}*)→

*L*. If , an open subset of

**R**, there must be an open ball . On the other hand, since (

*x*)→

_{n}*L*, some must lie within

*N*(

*L*, ε), which is a contradiction. ♦

Definition. We say that a subset is aclosed subsetif D-U is an open subset of D.

Warning. A subset of D can be both open and closed. E.g. the empty set (or the entire set D) is so.

**Useful note**: in particular, a closed and bounded subset *D* of **R** must have a maximum and minimum. Indeed, since *D* is bounded, let *M* be its supremum; we need to show that *M* lies in *D*. *Suppose it doesn’t*. For any positive ε, *M*-ε is not an upper bound of *D*, so there exists an element , *x* > *M*-ε and *x* < *M*. This shows that *M* is a point of accumulation of *D*, and by the above proposition, it must lie in *D*, which contradicts our assumption. Thus *M* is a maximum of *D*. The case for minimum is similar.

**Examples**

- Consider the set . It is closed in
**R**since the complement is which is a union of open intervals. Clearly*D*is bounded. Being both closed and bounded, it has a minimum and a maximum (in this case, 1 and 3 respectively). - Consider the set . It is not closed because its complement is and the point 2 has no open neighbourhood of
**R**which is wholly contained in*E*. Or, look at it another way, if it were closed, then since it’s obviously bounded it would contain a maximum and a minimum, but 2 is the supremum of*D*which is not contained in*D*itself. - Consider the set . The complement is which is a union of open intervals and is hence open. But it’s not upper-bounded so it doesn’t have a maximum.

From the above proposition, if

Dis a closed subset ofRandf:D→Ris continuous, then any convergent sequence inDgets mapped to a convergent sequence.

**Important Exercises**

- Prove that
*f*:*D*→**R**is continuous if and only if for each closed subset*C*of**R**, the set is closed in*D*. - Find a continuous function
*f*:**R**→**R**and a closed subset*C*of**R**such that*f*(*C*) is not closed in**R**. - Prove that a union of finitely many closed subsets of
*D*is still closed in*D*. - Prove that an intersection of (possibly infinitely many) closed subsets of
*D*is still closed in*D*. - Suppose .
- Prove that if
*C*is closed in*D*, then*B*:=*C*∩*E*is closed in*E*. - Prove that conversely, if
*B*is closed in*E*, then*B*=*C*∩*E*for some closed subset*C*of*D*.

- Prove that if

**Answers** [Highlight to read. Compare with the earlier case of open subsets. ]

- If
*C*is closed in**R**, then*D*–*f*^{-1}(*C*) =*f*^{-1}(*D*–*C*) is open in*D*iff*f*^{-1}(*C*) is closed in*D*. Then use the fact that*f*is continuous iff for each open subset*U*of**R**,*f*^{-1}(*U*) is open in*D*.

- Let
*f*be defined by*f*(*x*)=1/*x*for*x*>1 and*f*(*x*)=1 for*x*≤1. The set*C*=[1, ∞) is closed in**R**but*f*(*C*) = (0, 1] is not closed in**R**. - Follows from the fact that the intersection of finitely many open subsets of
*D*is also open in*D*. - Follows from the fact that the union of arbitrarily many open subsets of
*D*is also open in*D*. - For (A), if
*C*is closed in*D*, then*D*–*C*is open in*D*and*E*–*B*=*E*– (*C*∩*E*) = (*D*–*C*) ∩*E*is open in*E*by definition. Thus*B*is closed in*E*. For (B), since*E*–*B*is open in*E*, we can write*E*–*B*=*U*∩*E*for some open subset*U*of*D*. Then*B*=*E*– (*U*∩*E*) = (*D*–*U*) ∩*E*where*D*–*U*is closed in*D*.

## Uniform Continuity

Now let’s consider the case where *D* is not closed and (*x _{n}*) approaches an

*L*not in

*D*.

Definition. The function isuniformly continuous(on D) if:

- for any ε>0, there is a δ>0 such that when x, y in D satisfies |x-y|<δ, we have |f(x)-f(y)|<ε.

Pictorially:

Once again we note the difference between standard continuity and uniform continuity. In the former case:

- For every
*y*in*D*and ε>0, there exists δ>0 such that for all*x*in*D*satisfying |*x*–*y*|<δ, we have |*f*(*x*)-*f*(*y*)|<ε.

In the latter case:

- For every ε>0, there exists δ>0 such that for all
*x*,*y*in*D*satisfying |*x*–*y*|<δ, we have |*f*(*x*)-*f*(*y*)|<ε.

Using logical quantifiers, these can be succinctly expressed as:

and ,

where *P*(*x*, *y*, δ, ε) says “|*x*–*y*|<δ implies |*f*(*x*)-*f*(*y*)|<ε”.

**Examples of Uniform Continuity**

- Let
*f*(*x*)=*x*on any subset*S*of**R**. We claim that*f*is uniformly continuous. Indeed, for any ε>0, let δ=ε. Then whenever |*x*–*y*| < δ and*x*,*y*in*S*, we easily get |*f*(*x*)-*f*(*y*)| = |*x*–*y*| < δ = ε, which satisfies the definition for uniform continuity. - Let
*f*(*x*)=1/*x*on (0, ∞). We shall prove that*f*is not uniformly continuous. Negating the definition of uniform continuity, we need to find an ε>0 such that for any δ>0, we can find*x*,*y*>0 with |*x*–*y*|<δ but |1/*x*-1/*y*|≥ε. So let’s take ε=1; for any δ>0, pick*x=*δ and*y*= δ/(δ+1) <*x*. Then |*x*–*y*| =*x*–*y*< δ but |1/*x*-1/*y*|=1. - On the other hand,
*f*(*x*)=1/*x*on [1, 2] is uniformly continuous. Indeed, whenever 1 ≤*x*,*y*≤ 2, we have |1/*x*– 1/*y*| = |*x*–*y*|/|*xy*| ≤ |*x*–*y*| since |*xy*| ≥ 1. Hence, given any ε>0, we can just let δ=ε. Now whenever*x*,*y*in [1, 2] satisfy |*x*–*y*| < δ, we also have |*f*(*x*)-*f*(*y*)| ≤ |*x*–*y*| < δ = ε.

Examples 2 and 3 tell us that uniform continuity depends heavily on the domain of the function.

## Main Theorems

The first main result in this article is:

Theorem. If (x_{n}) is a Cauchy sequence in D and f : D →Ris a uniformly continuous function, then (f(x_{n})) is also a Cauchy sequence.

**Proof**. Let ε>0.

- Since
*f*is uniformly continuous, pick δ>0 such that when*x*,*y*in*D*satisfies |*x*–*y*|<δ, we have |*f*(*x*)-*f*(*y*)|<ε. - Since (
*x*) is Cauchy, pick an_{n}*N*such that when*m*,*n*>*N*, we have |*x*–_{m}*x*|<δ._{n}

This implies that when *m*, *n* > *N*, |*f*(*x _{m}*)-

*f*(

*x*)|<ε. ♦

_{n}For example, consider *f*(*x*)=1/*x* on (0, ∞). The sequence is Cauchy but the resulting is not a Cauchy sequence. Hence, this tells us *f*(*x*) is not uniformly continuous, as we had verified earlier directly in example 2 above.

Finally, let’s combine uniform convergence and uniform continuity.

Theorem. Suppose is a sequence of uniformly continuous functions on D, which converges uniformly to . Then is also uniformly continuous.

**Proof**.

Let ε>0. Then by uniformly convergence, there exists *N* such that whenever *n* > *N*, we have . Fix *n*. Since is uniformly continuous, there exists δ>0 such that whenever satisfies |*x*–*y*|<δ, we also have . Hence:

.

Now just go back and replace ε by ε/3. ♦