Question 1. Suppose that X is a nonempty set and T , T0 are topologies of X and B, B0 are bases of T , T0, respectively.
(1) For T1, T2 defined below, explain which one is a topology of X and which one is not.
T1 = {S ∈ P(X) : S ∈ T or S ∈ T0}, T2 = {S ∈ P(X) : S ∈ T and S ∈ T0}.
(2) For B1, B2 defined below, explain which one is a basis of a topology of X and which one is not.
B1 = {B ∈ P(X) : B ∈ B or B ∈ B0}, B2 = {B ∈ P(X) : B ∈ B and B ∈ B0}.
Answer: (1) T1 might not be a topology for X and T2 is a topology for X.
For T1, it is possible that S ∈ T \ T0 and S0 ∈ T0 \ T . Therefore, it is possible that S, S0 ∈ T1 but S ∪ S0 6∈ T and S ∪ S0 6∈ T0. For example, consider X = {1, 2, 3} and T = {X, {1, 2}, {3}, ∅}, T0 = {X, {1, 3}, {2}, ∅}. Then T1 = {X, {1, 2}, {1, 3}, {2}, {3}, ∅}.
Clearly, {2}, {3} ∈ T1, but {2} ∪ {3} = {2, 3} 6∈ T1. Therefore, T1 is not a topology for X.
For T2, if Ui ∈ T2, ∀ i ∈ I, then Ui ∈ T and Ui ∈ T0. Hence ∪i∈IUi ∈ T and ∪i∈IUi ∈ T0. This implies ∪i∈IUi ∈ T2. Moreover, if I is a finite set, then ∩i∈IUi ∈ T , then ∩i∈IUi ∈ T0. Hence ∩i∈IUi ∈ T2.
(2) Both B1 and B2 might not be a basis for a topology of X.
For B1, it is possible that B ∈ B \ B0 and B0 ∈ B0\ B. Therefore B, B0 ∈ B1, but it is possible that for some x ∈ B ∩ B0, there is no B00 containing x such that B00 ⊆ B ∩ B0 and B00 ∈ B1 (i.e. B00 ∈ B or B00 ∈ B0). For example, consider X, T , T0 as above. We have B = {{1, 2}, {3}} and B0 = {{1, 3}, {2}} are basis for T and T0, respectively. Therefore, B1 = {{1, 2}, {1, 3}, {2}, {3}}. However, 1 ∈ {1, 2} ∩ {1, 3}, but there is no element in B1 which is contained in {1, 2} ∩ {1, 3}.
For B2, it is possible that B2 is empty. For example, for B, B0 above, we have B2 = ∅. It
is clear that B2 is not a basis for any topology of X.
Question 2. Let X = {1, 2, 3, 4, 5} be a topological space with basis B = {{1, 2}, {3, 4}, {5}}.
Let ∼ be an equivalence relation on X, with equivalence class [1] = {1, 3}, [2] = {2, 4}, [5] = {5}. Denote the quotient space X/∼ = {1, 2, 5} with quotient space topology.
(1) Find all the open sets of X.
(2) Find all the open sets of X/∼.
(3) Is the quotient map q : X → X/∼ an open map?
Answer: (1) By definition the topology for X with basis B is
T = {X, {1, 2, 3, 4}, {1, 2, 5}, {3, 4, 5}, {1, 2}, {3, 4}, {5}, ∅}.
(2) Since the power set for {1, 2, 5} is
{{1, 2, 5}, {1, 2}, {1, 5}, {2, 5}, {1}, {2}, {5}, ∅},
we can check their inverse image of the quotient map q. Clearly {1, 2, 5} and ∅ are open.
Since q−1({1, 2}) = {1, 3, 2, 4} ∈ T , q−1({1, 5}) = {1, 3, 5} 6∈ T , q−1({2, 5}) = {2, 4, 5} 6∈ T , q−1({1}) = {1, 3} 6∈ T , q−1({2}) = {2, 4} 6∈ T and q−1({5}) = {5} ∈ T , we conclude that
T = {{1, 2, 5}, {1, 2}, {5}, ∅}˜ is the topology for X/∼.
(3) Because q({1, 2, 3, 4}) = q({1, 2}) = q({3, 4}) = {1, 2}, q({1, 2, 5}) = q({3, 4, 5}) = {1, 2, 5} and q({5}) = {5} are open in X/∼, we conclude that q is an open map. Question 3. Let X be a topological space with basis B. Let S be a subset of X. Prove that
ext(S) = [
{U ∈B | U ∩S=∅}
U.
1
2
Answer: By definition x ∈ ext(S) means there exist V open in X such that x ∈ V and V ∩ S = ∅. Since B is a basis, there exists U ∈ B such that x ∈ U and U ⊆ V . Therefore, U ∩ S ⊆ V ∩ U = ∅. This implies that x ∈ S
{U ∈B | U ∩S=∅}U. On the other hand, if x ∈ S
{U ∈B | U ∩S=∅}U , then there is U ∈ B with U ∩ S = ∅ such that x ∈ U . Since U open, we
have x ∈ ext(S).
Question 4. Let X be a topological space and S ⊆ X. Prove that S is dense in X if and only if bd(S) = cl(Sc)
Answer: By definition X is dense means that X = cl(S). Notice that from X = cl(S) ∪ ext(S) and cl(S) ∩ ext(S) = ∅, X is dense is equivalent to ext(S) = ∅. Therefore, by cl(Sc) = bd(S) ∪ ext(S), if ext(S) = ∅, then cl(Sc) = bd(S). On the other hand, because cl(Sc) = bd(S) ∪ ext(S) and bd(S) ∩ ext(S) = ∅, the assumption bd(S) = cl(Sc) implies
ext(S) = ∅, and hence X is dense.
Question 5. Let X be a topological space and let S be a nonempty subset of X. Consider cl(S) as a subspace of X.
(1) Suppose that U is a nonempty open subset of cl(S). Prove U ∩ S 6= ∅.
(2) Prove that if T is a disconnected open subset of cl(S), then T ∩ S is not connected.
Answer: (1) By definition, there exists an open set V of X such that U = V ∩ cl(S). Since U is not empty, there is an x ∈ U . In other words, x ∈ V and x ∈ cl(S). Therefore, by the definition of cl(S) and V being an open neighborhood of x, we conclude that V ∩ S 6= ∅.
Hence, U ∩ S = (V ∩ cl(S)) ∩ S = V ∩ S 6= ∅.
(2) By assumption, there exist U, U0 which are open in cl(S) such that T ⊆ U ∪ U0, U ∩ U0 = ∅, U ∩ T 6= ∅ and U0∩ T 6= ∅. Clearly, we have T ∩ S ⊆ U ∪ U0 and since U ∩ T and U0∩ T are nonempty open subset of cl(S), by (1), we have U ∩ (T ∩ S) = (U ∩ T ) ∩ S and U0∩ (T ∩ S) = (U0∩ T ) ∩ S are not empty. This shows that T ∩ S is disconnected. Question 6. Let X be a connected topological space and let ∼ be an equivalence relation on X. Suppose that X/∼ is a finite set with more than one element. Prove that as a quotient space, X/∼ is never a Hausdorff space.
Answer: Suppose on the contrary that X/∼ is a Hausdorff space. Which means that every one element set in X/∼ is closed, and hence by X/∼ is a finite set, we conclude that X/∼
is discrete. However, X being connected implies X/∼ is connected. This contradicts to the
assumption that X/∼ has more than one element.
Question 7. Let X be a Hausdorff topological space and Y be a compact topological space.
Suppose that f : X → Y is a bijective open function. Prove that f is continuous.
Answer: Since f : X → Y is a bijective open function, we have the inverse function f−1 : Y → X is continuous. Moreover, the assumptions Y is compact and X is Hausdorff imply that f−1 : Y → X is an open map, which is equivalent to f : X → Y is continuous. Question 8. Suppose that X is a Hausdorff topological space and S is a compact subset of X. Prove that for every x ∈ int(S), there exist an open set U such that x ∈ U and cl(U ) ⊆ int(S).
Answer: Since bd(S) is closed, the assumption S being compact implies that bd(S) is compact. For x ∈ int(S), since x 6∈ bd(S), by the assumption that X is a Hausdorff, there exist disjoint open sets U, V in X such that bd(S) ⊆ V and x ∈ U ⊆ int(S). Because U ∩ V = ∅ and X \ V is closed, by U ⊆ X \ V , we have cl(U ) ⊆ X \ V ⊆ X \ bd(S). This shows that cl(U ) ∩ bd(S) = ∅. Moreover, U ⊆ int(S) ⊆ S implies cl(U ) ⊆ cl(S) and hence
cl(U ) = cl(U ) ∩ cl(S) = cl(U ) ∩ (bd(S) ∪ int(S)) = cl(U ) ∩ int(S),
which implies that cl(U ) ⊆ int(S).
3
Question 9. Let X, Y be nonempty topological spaces such that X ∩ Y = ∅. Consider the disjoint union topological space X q Y and the direct product space X × Y . For fixed x0 ∈ X and y0 ∈ Y , consider the function f : X q Y → X × Y , defined by f (x) = (x, y0) and f (y) = (x0, y), ∀ x ∈ X, y ∈ Y . Consider Z = f (X q Y ) as a subspace of X × Y .
(1) Prove that f : X q Y → X × Y is a continuous function.
(2) Define g : X q Y → Z, by setting g(s) = f (s), ∀ s ∈ X q Y . Is f an open function?
Is g an open function? Show your reasoning.
(3) Define an equivalence relation ∼ on X q Y , by setting s ∼ s0 if and only if s = s0 or s, s0 ∈ {x0, y0}. Prove the quotient space (X q Y )/∼ is homeomorphic to Z.
Answer: (1) Since every open set of the basis of the product space X × Y is of the form U × V where U is open in X and V is open in Y , to show that f is continuous, we only need to check that f−1(U × V ) is open in X q Y . If x0 6∈ U and y0 6∈ V , then clearly f−1(U × V ) = ∅, which is open in X q Y . If x0 ∈ U and y0 6∈ V , then f−1(U × V ) = V which is open in Y and hence is open in X q Y . Similarly, if x0 6∈ U and y0 ∈ V , then f−1(U × V ) = U and if x0 ∈ U and y0 ∈ V , then f−1(U × V ) = U × V . For all these cases, we have that f−1(U × V ) is open in X q Y , and hence f : X q Y → X × Y is a continuous function.
(2) Neither f nor g is an open function. Suppose that U is open in X and {y0} is not open in Y . Then f (U ) = g(U ) = U × {y0}, which is not open in both X × Y and Z.
For example, consider X = {1, 2} and Y = {3, 4} both with indiscrete topology. Then the product topology for X × Y is also indiscrete. However X is open in X q Y , but for choosing x0 = 1, y0 = 3, we have f (X) = {(1, 3), (2, 3)} which is not open in X × Y . Since Z = f (X q Y ) = {(1, 3), (2, 3), (1, 4)}, considering as subspace of X × Y , Z is also indiscrete.
Hence f (X) is also not open in Z.
(3) Because the only case that s 6= s0 with s ∼ s0 is the case s = x0, s0 = y0, in this case we have g(s) = g(x0) = (x0, y0) = g(y0) = g(s0), and hence by the universal property for the quotient map q : X qY → (X qY )/∼, there exists a continuous function h : (X qY )/∼ → Z such that g = h ◦ q. Since g is onto, we also have h is onto. For showing h is one-to-one, we only have to show that if x ∈ X and y ∈ Y with h(x) = h(y), then x = y. In fact, h(x) = h(q(x)) = g(x) = (x, y0) and h(y) = h(q(y)) = g(y) = (x0, y). The assumption h(x) = h(y), implies x = x0, y = y0, and hence x = y. This shows that h is injective.
Finally, we need to show that h is an open map. Consider an open set W of (X q Y )/∼. If x0 6∈ W , then we have W = q(U ) ∪ q(V ), where U is open in X, V is open in Y and x0 6∈ U , y0 6∈ V . In this case, we have h(W ) = g(U ) ∪ g(V ) = (U × {y0}) ∪ ({x0} × V ). Because Z = (X × {y0}) ∪ ({x0} × Y ) and U ∩ {x0} = V ∩ {y0} = ∅, we have
((U ×Y )∪(X×V ))∩Z = ((U ×Y )∩(X×{y0}))∪((X×V )∩({x0}×Y )) = (U ×{y0})∪({x0}×V ).
In other words, h(W ) = ((U × Y ) ∪ (X × V )) ∩ Z. Since (U × Y ) ∪ (X × V ) is open in X × Y , we have h(W ) is open in Z. If x0 ∈ W , then we have W = q(U ) ∪ q(V ), where U is open in X, V is open in Y and x0 ∈ U , y0 ∈ V . In this case, we have
h(W ) = g(U ) ∪ g(V ) = (U × {y0}) ∪ ({x0} × V ) = (U × V ) ∩ Z.
Since U × V is open in X × Y , we have h(W ) is open in Z. It concludes that h is an open
function, and hence h is a homeomorphism.
