Suppose is a continuous function (where I is an interval). Now, for every and , we have a such that for . This tells us that f is bounded in the interval . Does it mean that the function is bounded in its entire domain? What we have shown is that given an , there is an interval and two real numbers and such that

for all .

Surely . But, if we could choose finitely many intervals out of the collection , say, such that , then we would get , where and . That, we can indeed make such a choice is a property of a closed bounded interval I in and is given by the following theorem, the proof of which, is given below:

**Theorem (Heine-Borel):**

Let and let I be a family of open intervals covering , that is, for all , there exists such that . Then, we can find finitely many open intervals such that .

**Proof:**

Suppose our contention is false: Let us take the intervals and where . If the hypothesis is false, then it should be false for at least one of the intervals or . Otherwise, we could find and such that and and then would be the finite family of intervals for which .

So let us assume that at least for one of the intervals or the assumption of the theorem is false. Call it . Again let . Now since the claim of the theorem is false for it should be false for at least or by the above argument. Call it . We have . We can continue this process to get a sequence of intervals for which the assertion is false. Observe further that and that we have .

This gives us a monotonically increasing sequence which is bounded above and a monotonically decreasing sequence bounded below. So and must converge to say and respectively. Then, because . Since covers , must belong to J for some . Also, since , there exists an such that for all . Now let . Therefore, we conclude that for all . But, this violates our hypothesis that we cannot choose finitely many members of whose union will contain for any n. QED.

**Corollary.**

A continuous function on a closed interval is bounded.

The proof of the corollary is already given just before the Heine-Borel theorem. So, if we have a continuous function and and , the above corollary says . Next, we ask the natural question: do there exist two points such that and In other words, does a continous function on a closed interval attain its bounds? The answer is yes.

**Theorem:**

Suppose is continuous, and and . Then, there are two points such that and .

Note: these points and need not be unique.

**Proof by contradiction:**

Suppose there is no point such that , then we would have or for all . Let us define by

Since vanishes nowhere, y is also a continuous function. So, by the corollary above it ought to be bounded above, and below. Let , for all . On the other hand, by the property of a supremum we note that there exists an such that , which implies that or , which is contradiction. Therefore, must attain the value M at some point . The proof of the other part is very similar. QED.

The above theorem together with the corollary says that on a closed interval, a continuous function is bounded and attains its bounds. This, again by the intermediate value theorem, means that the function must attain all the values between its supremum and infimum. Thus, the image of a closed interval under a continuous map is a closed interval. However, if f is a continuous map on an open interval, then the function need not be bounded.

**Example. **

Let be defined by . This is surely continuous but the limit,

, which means that given any , we can always find x such that , viz., choose .

If f is a continuous function, then given , for each fixed, we can find such that

whenever

Here depends upon .

Can we choose such that it works for all ? The answer in general is no.

**Example.**

Let be defined by . If we fix any , then for , , and hence as x becomes large, the difference between and also becomes large for every fixed . So for say , we cannot choose such that such that for all and all x. We thus have the following definition:

**Definition:**

Let be a continuous function where or or (a,b). Then, f is said to be uniformly continuous if for all , there exists a such that

for all with

We have seen above that every continuous function need not uniformly continuous. When , however, every continuous function is uniformly continuous as the next result shows.

**Theorem:**

Let be continuous. Then, f is uniformly continuous.

**Proof.**

Fix . The continuity of f implies that for every , we can choose such that

whenever and

Now, let

Then, clearly covers as . By the Heine Borel theorem, we can get finitely many intervals out of this family, , , …, such that

.

Let

Then, (note that minimum of finitely many positive numbers is always positive). Next we claim that if , then

Since , we can find such that , that is, . Now, .

Hence, and and therefore, . QED.

More later,

Nalin Pithwa