MAT 364 Topology
Problem Set 3
2.19. Show that every set A is both open and closed relative to itself.
Proof. The question is almost trivial if you understand what it says. There
are several ways to solve it. Let’s look carefully.
OPEN: by definition, we need to show that every point of A is interior
relative to A, i.e. that every x ∈ A has a relative neighborhood entirely
contained in A. But a relative neighborhood is a set of the form Nx =
D(x, r) ∩ A; because we intersect with A, any relative neighborhood of any
point is always contained in A.
CLOSED: by definition, we need to check that every point in the complement
of A in A is exterior. But A − A = ∅, so there are no points for which we’d
have to check the exterior condition. Logic tells us that this condition is
Another solution would be to remember that relatively open sets in A are
exactly sets of the form A ∩ O, where O open in Rn , and similarly relatively
closed sets in A are of the form A ∩ C, C closed in Rn (Thm 2.14). Since
A = A ∩ Rn , and Rn is both open and closed, the statement follows.
2.20. Show that the empty set is both open and closed relative in any A.
Proof. This is similar to 2.19 – in fact can be derived from 2.19 using 2.22
(how?). Again, a proof from neighborhoods and a proof from 2.14 are possible. The proof from 2.14 uses the fact that ∅ = A ∩ ∅, and the empty set
is both open and closed in Rn . The neighborhoods proof is below.
OPEN: need to check that every point in the empty set is interior rel A.
There are no points, thus nothing to check.
CLOSED: need to check that every point in A − ∅ = A is exterior relative to
A. This means that every point x ∈ A comes with a relative neighborhood
Nx contained in A. But as in 2.19, every Nx is contained in A since Nx =
D(x, r) ∩ A.
2.22. A open relative to X iff X − A closed rel to X.
Proof. Again we can argue from neighborhoods or from 2.14. Using 2.14 is
perhaps quicker: A open rel X ⇐⇒ A = X ∩ O for some O open in Rn
⇐⇒ (∗) X − A = X ∩ C for some C closed in Rn ⇐⇒ X − A closed in X.
To see why (∗) is true, set C = Rn − O, and use the fact that C closed iff O
open (in Rn !).
2.25. A composition of two continuous functions is continuous.
Proof. One can argue from −δ definition, but open sets give a neater proof.
Consider continuous functions
f : A → B,
Let U be an arbitrary open set in C, then V = g−1 (C) is open in B. But
then, since f is continuous, f −1 (V ) = f −1 (g−1 (C)) = (g ◦ f )−1 (U ) is open.
It follows that (g ◦ f ) is continuous.