ASSIGNMENT 2 SOLUTIONS
TOPOLOGY BASES
IGOR WIGMAN
PLEASE INFORM ME IF THERE ARE ANY INQUIRIES/MISPRINTS.
(1) We choose to use Lemma 2 (say) from class. First, every such interval (a, b) with a, b ∈ Q
is open (in the standard topology). It is then sufficient to show that for every open U ⊆ R
(in the standard topology) and every x ∈ U there exists an interval I = (a, b) with
a, b ∈ Q so that x ∈ I ⊆ U . Since U is open, there exists > 0 s.t. (x − , x + ) ⊆ U . By
the density of rational numbers in R, there exists a ∈ Q ∩ (x − , x) and b ∈ Q ∩ (x, x + ),
and it is easy to see that I = (a, b) satisfies the condition. That every open set U ⊆ R
is a countable union of open intervals follows from Lemma 1 (why?).
(2) (a) Proof of (i) ⇒ (ii): Here it is sufficient to notice that if B ∈ B then, in particular,
B ∈ T ⊆ T 0 is open in topology generated by B 0 . The existence of B 0 ∈ B 0 with
x ∈ B 0 ⊆ B then follows from the definition of T 0 being generated by B 0 .
Proof of (ii) ⇒ (i): Let U ∈ T , we would like to show U ∈ T 0 . By the definition of
topology T generated by B, it follows that for every x ∈ U , there exists B ∈ B, such
that x ∈ B ⊆ U . Then, by (ii), there exists B 0 ∈ B0 such that x ∈ B 0 ⊆ B ⊂ U , so
given x ∈ U , we found B 0 ∈ B 0 so that x ∈ B 0 ⊆ U , and thus U is open in T 0 , by
the definition of the fact that T 0 is generated by B 0 .
(b) (For every x ∈ X and B ∈ B such that x ∈ B, there exists B 0 ∈ B0 such that
x ∈ B 0 ⊆ B) and (For every x ∈ X and B 0 ∈ B 0 such that x ∈ B 0 , there exists
B ∈ B such that x ∈ B ⊆ B 0 ).
(c) Here it is sufficient to impose the condition ”for every x ∈ X and B 0 ∈ B 0 such that
x ∈ B 0 , there exists B ∈ B such that x ∈ B ⊆ B 0 ”, the other one being satisfied
automatically. What we needed to check for Q1 is precisely that for every open
interval (c, d) with c, d ∈ R and x ∈ (c, d) there exist rational a, b ∈ Q so that
x ∈ (a, b) ⊆ (c, d), which is what we did.
(3) An intersection of two squares is not necessarily a square, but it is a rectangle, and it is
possible to insert a square inside it, around any arbitrary point x in the intersection. The
family B is a subset of rectangles (which are known to generate the standard topology
on R2 from lectures); it is then sufficient that the topology generated by B is finer than
the topology generated by rectangles, like in Q2. To this end we note that inside every
rectangle, it is possible to insert a square, around any given point. It would be better to
add a picture as an illustration for each of the above assertions to get full marks. Same
story works for ellipses and higher dimensions (replacing circles by balls, rectangles by
boxes, squares by cubes and ellipses by ellipsoids).
(4) The topologies T3 and T4 , that are not comparable between themselves. Both T3 and T4
are strictly finer than the standard topology T1 , which is already known to be strictly
finer than the finite complement topology T2 . To see this, just use Q2. It is better to
draw a tree of comparison between topologies to illustrate the answer.
(5) The collection
{[a, b] : a < b}
1
2
IGOR WIGMAN PLEASE INFORM ME IF THERE ARE ANY INQUIRIES/MISPRINTS.
is not a topology basis since it fails to satisfy the second axiom of a basis ([a, b] ∩ [b, c] =
{b} which forces all the singletons to be contained in the basis - why?). The collection
{[a, b] : a ≤ b}
is a topology basis, and it generates the discrete topology since it contains all the singletons.
(6) Let
\
T0=
Ti ,
i∈I
where the intersection is over all topologies containing B. Then T 0 is a topology by
Assignment 1, Q2; B ⊆ T 0 since ∀i ∈ I.B ⊆ Ti . It is then obvious that T 0 is the coarsest
topology containing B (since it is the intersection of such), and then T 0 ⊆ T . On the
other hand, any topology Ti containing B would also contain arbitrary unions of sets in
B, and thus, by Lemma 1, also B. It is then evident that T ⊆ T 0 .
(7) Let U ⊆ R be a nonempty clopen set and let x0 ∈ U . We claim that necessarily
y0 = y0 (x0 ) = ∞. If x0 < y0 < ∞, then [x0 , y0 ) ⊆ U . Now, if y0 ∈ U , then [x0 , y0 ] ⊆ U ,
and for every epsilon
(y0 , y0 + ) \ U 6= ∅.
This is a contradiction to openness of U . If otherwise, y0 ∈
/ U , then y0 ∈ R \ U , which is
a contradiction of openness of R\U (closedness of U ). Thus y0 = ∞, so that [x, ∞) ⊆ U ,
and similarly, (−∞, x] ⊆ U , so that U = R.