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Basic Topology - M.A.Armstrong
Answers and Solutions to Problems and Exercises
Gaps (things left to the reader)
and Study Guide
1987/2010 editions
Gregory R. Grant
University of Pennsylvania
email: [email protected]
April 2015
A Note to the (Potential) Reader of M.A. Armstrong
I think this is a great book. But from a user’s perspective, I’m afraid a lot of people never get through chapter one. At
issue are Armstrong’s seemingly casual approach and his initial and confusing defnition of topology using “neighborhoods”.
Doing things that way gets confusing real fast, particularly (1.3) (d) on page 13. I imagine some readers are turned away at
this point, frustrated by the apparent complexity of the subject.
But happily things become much simpler and clearer starting in chapter 2. Chapter 2 basically starts over, doing things
the “right” way (as opposed to the “intuitive” way of chapter 1). The author’s intention was obviously to introduce the subject
by somewhat informally developing some motivation and intuition. And if you’re able to overcome the initial confusions
relating to “neighborhoods”, then he definitely does achieve a decent overview, which serves as a nice starting point for
studying topology. The reader should be patient and try to get the general concepts without buckling under the weight of the
details of (1.3).
This won’t cost you anything, because for the most part everything starts back over in chapter 2 where the concept of a
“topology” is defined anew in much simpler terms. There are a few exercises to show the two approaches are equivalent (you
can read their solutions here in these notes), but they’re not that critical. Just read chapter one as best you can, after reading a
few chapters you can go back and review chapter 1 and it will make a lot more sense.
If you find any mistakes in these notes, please do let me know at one of these email addresses:
[email protected] or [email protected] or [email protected]
Chapter 1 - Introduction
Page 8. Armstrong presents a formula for the area of Q as α1 + α2 + · · · + αk − (n − 2)π as if it is obvious. This is NOT actually
supposed to be obvious.
Page 14, Example 2. He says the axioms for a topology are easily checked. This is Chapter 1, Problem 19.
Page 14, Example 3. It is stated that the ”inverse is not continuous” and we are asked ”Why not?” This is Chapter 1, Problem
Page 14, Example 4. We are asked to ”Check that radial projection π gives a homeomorphism” in Fig. 1.8. This is Chapter 1,
Problem 20.
Page 14, Example 6. We are asked to why the finite complement topology on R cannot be given by a metric. Suppose R (with
finite complement topology) is a metric space with metric d. Let x, y ∈ R and let D = d(x, y) be the distance between x and y.
Let Bx be the set of points that are strictly less than D/2 units ditsance from x and let By be the set of points that are strictly
less than D/2 units ditsance from y. (All distances mentioned are with respect to d, not to the typical distance in R.) Then
Bx and By must be non-empty and disjoint. But in the finite complement topology no two non-empty open sets can be disjoint.
Problem 1. Prove that v(T ) − e(T ) = 1 for any tree T .
Solution: Any tree can be obtained by starting with a single edge and then attaching edges one at a time so that the graph is
connected at each step. We will denote a partial tree by T 0 . Since it is a tree, each attachment adds one edge and one vertex.
Therefore each attachment after the first does not change v(T 0 ) − e(T 0 ) (where T 0 is the partial tree as built so far). Only the
first edge is different. The first edge introduces one edge and two vertices. Therefore at step one v(T 0 ) − e(T 0 ) = 1 and adding
each additional edge does not change it further. Therefore v(T ) − e(T ) = 1 for the whole tree.
Problem 2. Even better, show that v(Γ) − e(Γ) ≤ 1 for an graph Γ, with equality precisely when Γ is a tree.
Solution: We assume the graph is connected, otherwise it is obviously false. In the same way as before we can build the graph
edge by edge. The first edge gives v(T 0 ) − e(T 0 ) = 1. Since we add edges so that T 0 is always connected, each additional edge
adds exactly one edge and at most one additional vertex. Therefore v(T 0 ) − e(T 0 ) is adjusted at each step by either 0 or −1.
So the final sum v(T ) − e(T ) cannot be greater than one.
If the graph is not a tree then some edge must connect two existing vertices as we build T . Therefore at some step we will
add −1 and the final sum can not be greater than zero.
Problem 3. Show that inside any graph we can always find a tree which contains all the vertices.
Solution: If a graph has a cycle, then any edge in the cycle can be removed without causing it to become disconnected. To
see this suppose an edge in the cycle connects vertices A and B. Suppose C and D are any two nodes. Connect them by a
path in the original graph. Now if that path passes through the edge connecting A and B, it can be diverted around the cycle to
get from A to B the long way. This path is still valid in the graph with the edge connecting AB removed. Therefore there is a
path connecting C and D in the modified graph. So the graph remains connected. Now as long as there are cycles, continue to
remove edges until there are no more cycles. The remaining graph is connected with no cycles and therefore must be a tree.
Problem 4. Find a tree in the polyhedron of Fig. 1.3 which contains all the vertices. Construct the dual graph Γ and sohw
that Γ contains loops.
The tree that hits every vertex is shown in red.
The dual graph Γ is shown in blue with the loops being shown in green. There are two loops that connect at a point.
Problem 5. Having done Problem 4, thicken both T and Γ in the polyhedron. T is a tree, so thickening it gives a disc. What
do you obtain when you thicken Γ?
Solution: Γ is basically two loops connected at a point, with some other edges connected that do not make any more loops.
So thickening should procduce something homeomorphic to what is shown in Problem 11 (b) (right). What is called ”Two
cylinders glued together over a square patch”.
Problem 6. Let P be a regular polyhedron in which each face has p edges and for which q faces meet at each vertex. Using
Euler’s formula prove that
1 1 1 1
+ = + .
p q 2 e
Solution: Every edge in a polyhedron has two faces. Each face has p edges so f p is twice the total number of edges. So
f p = 2e. Therefore f = 2ep . Now each face has p vertices and each vertex is in q faces. So v = pqf . Euler’s formula says
v − e + f − 2. Therefore
= 2.
Divide through by 2e to get the result.
Problem 7. Deduce from Problem 6 that there are only five regular polyhedra.
Solution: Clearly p ≥ 3. Suppose p ≥ 6. Then
1 1 1 1 1 1
+ = + ≤ + .
2 e
p q 6 q
1 2 1 1
< + ≤ .
3 6 e q