ECON6003/6703 Mathematical Methods for Economics Problem Set 1 (Solutions)

Question 1. Starting from page 20 of your Lecture 1 slides, a number of environ- ments are described. In each of the questions, you are asked to identify certain sets.

State for each case, whether those sets are open, closed or compact.       Answer. For a video solution of this question click this link: Click Here

Question 2.  Given a pair of closed sets A ; B  Rm  which of the following sets are closed?  In each case, give a proof if you claim the set is closed, and a counter example otherwise.

1   A \ B           2   A n B           3   AB

Answer.

1)    A\B is a closed set.  We know from Theorem 3, that an arbitrary intersection of closed sets is closed. We proved part of that theorem in class, and the rest of it appears as solution to Question 5. Here let us try a direct proof that does not make any reference to open sets but only uses the fact that a set is closed if and only if it contains its boundary, i.e. Theorem 1.

Sample Proof.  From Theorem 1, Lec 2, we know that @(A)A and @(B)  B . We will use this and complete the proof by showing @(A\B)  A \ B .

Choose any  x 2 Ac . Then, x  @(Ac) since @(A)  A. That is to say, there exists some " > 0 such that either  B" (x)  A or B" (x)  Ac .   (In plain language, x is not a boundary point and therefore it must have a neighbourhood that lies entirely in A or entirely in Ac). But since x 2 Ac , it must be that B" (x)  Ac , which in turn implies that B" (x)(A\B)c . In other words, we exhibited an "-ball at x that lies entirely in (A\B)c , which means x  @(A\B).

We have shown that x  A ) x  @(A \ B), i.e.  Ac   (@(A \ B))c , which is the contrapositive of @(A \ B)  A.   A symmetric argument gives that @(A \ B)  B , i.e. @(A\B)  A\ B .

2).  We cannot conclude anything about A n B .  This is essentially A \ Bc .  It is the intersection of closed set with an open set. The result could be open, closed or neither open nor closed. Fool around with A = [a ; b] and B = [c ; d] for different values of a ; b ; c ; d to convince yourself.  Indeed, if you tke A=[0; 1] and B =f0; 1g, you get right away that A nB =(0; 1), whereas if you took B =f4g, you would get A n B = [0; 1] etc. You should draw a few pictures like the ones below to convince yourself of examples in R2 .

A nB

Figure 1.  A and B are areas surrounded by the rectangle and the triangle respectively, including the boundary. A nB is not closed (or open).

3) Similar to previous part.

Question 3.  Repeat the above question where the closedness is replaced with compactness .

Answer. No change to answer in previous question.

Question 4.  A set  is said to be a singleton if it has just one element. Pick any x 2 Rm and consider the singleton fxg.  Prove that fxg is a closed set. Next show that for any given finitely many points in x0 ; x1 ; ::: ; xk 2Rm , finite set S =fx1 ; ::: ; xk g is a closed set. Hint: You can use the various theorems on closed and open sets in Lec 2.

Answer.  Let X = fxg.  We will show that @(X) = X , (i.e.  that X contains its boundary).  By Theorem 1, we can conclude that X is closed.

Let us begin by showing x 2 @(X). Indeed, it is clear that for every ">0, B" (x)\Xc  ;, and since x 2 X , B" (x) \ X also holds. Therefore x is a boundary point of X . Next, we need to show that if y X , then it cannot be a boundary point of X . For this, pick any y  X , and let " = d(x ; y).  Since x  y , d(x ; y) > 0 and hence " > 0. Moreover, d(x ; y) =2" > ".  Therefore, x  B" (y), which means B" (y)\X =;. That is, we have exhibited a neighbourhood of y which does not intersect X , which is to say y  @(X).  Thus we conclude that @(X) =X .

Finally, showing S =fx1 ; ::: ; xk g is a closed set is easy. From the above arguments, we know that each of the sets fx1g; fx2g; ::: ; fxk g are closed. Since S is the union of these sets, it is a closed set, by Theorem 3.

Question  5.  Using Theorem 2 Lec 2 notes  (Theorem 12.8 from text) and the general results for set operations on page 19 of Lec 1 notes, give a complete proof of Theorem 3, Lec 2 notes.

Answer.

Sample Proof of Part 2, namely that  Finite union of closed sets is closed.

This requires an application of De Morgan's Laws (Proposition 3, Week 1 slides).

Let S1 ; S2 ; ::: ; Sn be a finite collection of closed sets. We need to show that S1 [:::[Sn is closed.

Since Sk  is closed, by definition,  Sk(c)     is open, for each k = 1; : : : ; n.