# Mean value theorem

Prove: If f : U → R is continuous on U, and E ⊂ U is closed and bounded, then
f attains an absolute minimum and maximum on E.

I have no idea even where to start on this. Intuitively it's so obvious that i don't know what to do. The definitions given by the teachers that I have to work with are as follows:

A set F ⊆ Rn is closed if, for every convergent sequence {xi}(from i=1 to infinity) ⊆ F,
we have limxn(as n goes to infinity)⊆F
(in other words it contains its limit points)

A bounded set is one for which there exists r such that the set is contained in Nr(0).
(in other words some ball around the origin of any size contains the set.)

F continuous on u means for all c in F, lim(as x approaches c) exists and equals F(c).

A compact set is one which is closed and bounded, so E in this proof is compact.

I know that what needs to be shown is that there exists xm and xM such that:
f(xm)<=f(x)<=f(xM) for all x in E.

any advice? hints? etc.. on how to start this? I suspect that the mean value theorem might have something to do with it, but I'm not sure how to incorporate it. Thanks for any help.

Galileo
Homework Helper
Hint: Consider the supremum of f(E).

matt grime
Homework Helper
Depending on your familiarity with these ideas:

Let K be the image of E, under F. Let C be a covering of K by open sets. Consider the inverse image of the sets in C by f. What do you know about compact sets? Can you use this here, what about mapping back to K? Is K thus compact?

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Unfortunately im not familiar with most of these terms. What's the supremum? What does a covering of K by open sets mean? I'm pretty sure I get the image part. Does the image of E under F simply mean the range of F if the domain were restriced to E?

eddo said:
Unfortunately im not familiar with most of these terms. What's the supremum? What does a covering of K by open sets mean? I'm pretty sure I get the image part. Does the image of E under F simply mean the range of F if the domain were restriced to E?
An upper bound for a set F (subset of R) is a real number M such that for all x in F, x <= M.

Example: Let F=(-oo,3]. Then 666 is an upper bound for F, as are 3 and 5 and any number not less than 3.

The supremum of a set F (subset of R) is the least upper bound of F.

Example: Let F=(-oo,3]. Then 3 is the supremum of F. 3 is also the supremum of (-oo,3).

A set G is open if for all x in G, there is an interval (x-d, x+d) fully contained in G. In other words, a d>0 such that (x-d,x+d) is a subset of G.

A covering of K by open sets means a family of sets, W, such that each member of W is open and K is a subset of the union of all sets in W. (The union of all sets in W is the set of all points in some element of W.)

For example, Let W={(-oo,n):n &isin; Z}. W is an open covering of the set K=[36,666].

The image of E under f, denoted f(E), is the set of elements in the range mapped to by some element of E. So y is in f(E) iff f(x)=y for some x in E.

matt grime