[From the 4AM Lectures series]

Today, my subject will be -dimensional Euclidean space, so everything I say here applies strictly in this context. Any respectable analysis course will certainly have you acquainted with some elementary point-set topology. We’ll review some important notions. Note that denotes the norm of .

**Definitions**. For a point we define the *open ball* of radius centered at by , and the *closed ball* of radius centered at by .

**More definitions**. Let . A point is said to be an *interior point* of if there exists such that the open ball is contained within . Now, we call a set *open* if every point in is an interior point. Furthermore, we call a set *closed* if the set (that is, the set of all points *not* in ) is open.

**Even more definitions**. A set is called *bounded* if there exists some open ball centered at that contains all of . Finally, a set is called *compact* if it is both closed and bounded. (The true notion of topological compactness is much more general, having to do with covering of sets. However, in the special case of , it is shown to be equivalent to the definition I have given here, by a rather significant result called the *Heine-Borel Theorem*.)

**Exercise**. Verify that open balls are indeed open sets and closed balls are indeed closed sets.

Now that we’re done with the modest topological groundwork we’ll need, we proceed to define what it means for a function to be continuous. Intuitively you may think of continuous functions as those functions which have no “jumps” in their graphs. However, this notion quickly shows itself insufficient for anything even remotely subtle. Also, the concept of a “jump” in the graph of a multivariable function becomes slightly vague.

**Definition**. Let be nonempty, and let . We say that a function is *continuous* at if for every there exists some such that whenever and , we have . Note that the first norm is a norm in , whereas the second is a norm in . If is continuous at every point in we often say that is continuous on .

So as it turns out, continuous functions and compact sets are pretty closely related. Related in the sense that something about the way these concepts are defined yields some nice interplay between them. Here are a few theorems.

**Theorem**. Let be compact, and let be a continuous function. Then the image set, , is compact.

Or, stated more elegantly, that the continuous image of a compact set is compact.

The importance of this result cannot be overstated. Why? Because a major result known as the *Extreme Value Theorem* is an immediate corollary. The Extreme Value Theorem says that any map attains its maximum and minimum value on , given that is compact! But if is compact, then by the theorem above, is compact, and certainly any compact subset of the real line contains its supremum and infimum! And hence by taking the preimages of the L.U.B. and G.L.B. we see the EVT falls out in this manner. Now we move on to another concept, namely that of uniform continuity. Take care in noting the difference between uniform continuity and continuity, which may at first seem subtle.

**Definition**. Let be nonempty, and let . We say that a function is *uniformly continuous* on if for every there exists some such that whenever with , we obtain .

If a function is uniformly continuous on a set, it is continuous at every point in that set. This is trivial to verify. We will now present something that is not so trivial to verify. But first, we should remark the following:

**Proposition**. If is nonempty, and is continuous, then whenever is a sequence in converging to some limit , we have that the sequence converges to .

More elegantly put, this simply says that for a continuous function, the limit of the images is the image of the limit. This is not hard to prove. Now we can comfortably prove our theorem:

**Theorem**. Let be a compact set, and let be continuous on . Then is uniformly continuous on .

**Proof**. Note that if is not uniformly continuous, there exists some such that no exists satisfying the definition. Fix , and assume, for the sake of contradiction, that indeed no such exists. For each natural number , put . Then our assumption furnishes points such that , but . We construct sequences and from these points. Since is compact, every sequence in has a subsequence that converges to a point that is again in , say , with , where . We claim that also. To see this, note that by the triangle inequality, .

But the rightmost expression here certainly converges to 0 as we let , and hence by the squeeze theorem, we obtain as previously claimed. Now, since the function is continuous on , it respects sequences converging to points within . Hence, the fact that implies that and similarly we obtain that . Then certainly as we obtain that . But this is a contradiction, since we have for each natural that ! Hence our assumption that was not uniformly continuous must have been false. Therefore, is uniformly continuous on as required.

Note that the whole mechanics of the above proof rely on two key parts: firstly, the compactness of the domain allows us to invoke the Bolzano-Weierstrass theorem, and select a subsequence that converges *within* . Secondly, the continuity of implies that the limit of the images is the image of the limit, which is where our contradiction comes from!

For more, see my MATH 247 notes.