p91: 4, 6 4.) Let W be open in X × Y. Show π_{1}(W) is open in X. Let x_{0} ∈ π_{1}(W). Find U open in X such that x_{0} ∈ U ⊂ π_{1}(W). x_{0} ∈ π_{1}(W) implies there exists an (x_{0}, y_{0}) ∈ W such that π_{1}(x_{0}, y_{0}) = x_{0}. W open implies there exists a basis element U × V where U is open in X and V is open in Y and (x_{0}, y_{0}) ∈ U × V ⊂ W Thus π_{1}((x_{0}, y_{0})) ∈ π_{1}(U × V) ⊂ π_{1}(W). Hence x_{0} ∈ U ⊂ π_{1}(W). 6.) R^{2} has the product topology. Thus by thm 15.1, {B_{1} × B_{2} | B_{1} ∈ B_{1}, B_{2} ∈ B_{2}} is basis for R^{2} where B_{i} are bases for R. By problem #8, p. 83, B = {(a, b) | a < b, a and b rational } is a basis for R. Thus {(a, b) × (c, d) | a < b and c < d, a, b, c, d rational } is a basis for R^{2}. p91: 1, 3, 8 1.) Let T_{X} be the topology on X. Let T_{Y} = {Y ∩ U | U ∈ T_{X}} Let T_{A} = {A ∩ U | U ∈ T_{X}} Let T′_{A} = {A ∩ U | U ∈ T_{Y} } Show T_{A} = T′_{A} Let W ∈ T_{A}. Then there exists a U ∈ T_{X} such that W = A ∩ U. U ∈ T_{X} implies Y ∩ U ∈ T_{Y}. Since A ⊂ Y, A ∩ (Y ∩ U) = (A ∩ Y) ∩ U = A ∩ U = W. Thus W ∈ T′_{A}. Hence T_{A} ⊂ T′_{A} Let W ∈ T′_{A}. Then there exists a U ∈ T_{Y} such that W = A ∩ U. U ∈ T_{Y} implies there exists a V ∈ T_{X} such that U = Y ∩ V. Since A ⊂ Y, W = A ∩ U = A ∩ Y ∩ V = A ∩ V. Thus W ∈ T_{A}. Hence T′_{A} ⊂ T_{A} Therefore T_{A} = T′_{A} 3.) A and E are open in R. B, C, D are not open in R. A, B, E are open in [−1, 1]. C, D are not open in [−1, 1]. A = (−1, −\frac{1}{2}) ∪ (\frac{1}{2}, 1) is open in R since A is the union of two basis element. B = [−1, −\frac{1}{2}) ∪ (\frac{1}{2}, 1]. Show B is not open in R. Show that there exists an x ∈ B such that for all basis elements, B_{x} such that x ∈ B_{x}, B_{x} ⊄ B. Let x = 1. Suppose 1 ∈ (a, b). Thus a < 1 < b. Hence a < 1 < \frac{1+b}{2} < b. Thus \frac{1+b}{2} ∈ (a, b), but \frac{1+b}{2} ∉ [−1, −\frac{1}{2}) ∪ (\frac{1}{2}, 1]. Thus, (a, b) ⊄ B. Thus B is not open. Similarly C (take x = \frac{1}{2}) and D (take x = \frac{1}{2} or 1) are not open.

Take x ∈ E = \{x | 0 < |x| < 1 and \frac{1}{x} ∉ Z_{+}\}. If x < 0, then x ∈ (−1, 0) ⊂ E. If x ≮ 0. Then x ≠ \frac{1}{n} for any n ∈ Z_{+} and x ≠ 0 since x ∈ E. Let F = \{n ∈ Z_{+} | \frac{1}{n} < x\}. Since F is bounded below by 0, inf F exists in Z since Z has the greatest lower bound property. Let N = inf F. Then \frac{1}{N} < x. N − 1 < N implies that N − 1 ∉ F. Since x < 1, N ≠ 1, and thus N − 1 ∈ Z_{+}. Hence \frac{1}{N−1} ≥ x and since x ≠ \frac{1}{N−1}, \frac{1}{N−1} > x. Thus, x ∈ (\frac{1}{N} , \frac{1}{N−1}) ⊂ E. Thus E is open in R.

Alternate proof that E is open in R. E = (−1, 0) ∪ [∪_{n=1}^{∞}(\frac{1}{n+1}, \frac{1}{n})]. Thus E is open since it is the union of open sets.

Question: Can you prove that E = (−1, 0) ∪ [∪_{n=1}^{∞}(\frac{1}{n+1}, \frac{1}{n})].

A open in R and A = A ∩ [−1, 1] implies A is open in [−1, 1].

(−3, −\frac{1}{2}) ∪ (\frac{1}{2}, 5) open in R and B = [(−3, −\frac{1}{2}) ∪ (\frac{1}{2}, 5)] ∩ [−1.1] implies B is open in [−1.1].

C = (−1, −\frac{1}{2}] ∪ [\frac{1}{2}, 1) Show C is not open. Show that there exists an x ∈ C such that for all basis elements, B_{x} such that x ∈ B_{x}, B_{x} ⊄ C.

Let x = \frac{1}{2}. Suppose \frac{1}{2} ∈ (a, b) ∩ [−1, 1]. Let a′ = max\{a, −\frac{1}{2}\}. Thus a ≤ a′ < \frac{1}{2} < b. Hence

a < \frac{\frac{1}{2} +a′}{2} < \frac{1}{2} < b.

Also −\frac{1}{2} ≤ a′ < \frac{1}{2} implies −1 < −\frac{1}{2} < \frac{\frac{1}{2} +a′}{2} < \frac{1}{2} < 1 Thus \frac{\frac{1}{2} +a′}{2} ∈ (a, b) ∩ [−1, 1], but \frac{\frac{1}{2} +a′}{2} ∉ (−1, −\frac{1}{2}] ∪ [\frac{1}{2}, 1). Thus, (a, b) ∩ [−1, 1] ⊄ C. Thus C is not open.

Similarly D is not open.

E open in R and E = E ∩ [−1, 1] implies E is open in [−1, 1]

8.) Let R^{2} have the topology generated by T_{X} × T_{Y}.

Let L be a line with slope 0. Then L = R × {y_{0}} for some y_{0} ∈ R. Note that L = R × {y_{0}} has the subspace topology by hypothesis. By thm 16.3, the subspace topology on L = R×{y_{0}} as a subspace of R^{2} is the same as the product topology on R×{y_{0}} where R is a subspace of (R, T_{X}) and {y_{0}} is a subspace of (R, T_{Y}). Thus, R×{y_{0}} has the topology generated by {U ×V | U ∈ T_{X}, V ∈ {∅, {y_{0}}}}. Since A × ∅ = ∅, {U × V | U ∈ T_{X}, V ∈ {∅, {y_{0}}}} = {U × V | U ∈ T_{X}, V = {y_{0}}}. Thus the topology on L is equivalent to T_{X}. Thus L inherits the lower limit topology as a subspace of R_{l} × R and L inherits the lower limit topology as a subspace of R_{l} × R_{l}.

Similarly if L is a line with vertical slope, the the topology on L = T_{Y}. Thus L inherits the standard topology as a subspace of R_{l} × R and L inherits the lower limit topology as a subspace of R_{l} × R_{l}.

Let R^{2} have the topology generated by R × R_{l}.

Let L be a line with positive slope. Let f(x) = mx + b where m > 0. L = {(x, f(x)) | x ∈ R}. (a, b) × [c, d) is a basis element for the topology generated by R × R_{l}.

If f(a) ≥ d or f(b) ≤ c, then

{(x, f(x)) | x ∈ R} ∩ (a, b) × [c, d) = ∅ If f(a) < c and f(b) ≥ d then {(x, f(x)) | x ∈ R} ∩ (a, b) × [c, d) = [(f^{−1}(c), c), (f^{−1}(d), d)) where [(f^{−1}(c), c), (f^{−1}(d), d)) = {(x, f(x)) | f^{−1}(c) ≤ x < f^{−1}(d)} If f(a) < c, c < f(b) ≤ d then {(x, f(x)) | x ∈ R}∩(a, b)×[c, d) = [(f^{−1}(c), c), (b, f(b))) where [(f^{−1}(c), c), (b, f(b))) = {(x, f(x)) | f^{−1}(c) ≤ x < b, } Thus since f is onto, every half open interval in L is open. Thus the topology on L is equal to or finer that the lower limit topology. If c ≤ f(a) < d and f(b) ≥ d then {(x, f(x)) | x ∈ R}∩(a, b)×[c, d) = ((a, f(a)), (f^{−1}(d), d)) where ((a, f(a)), (f^{−1}(d), d)) = {(x, f(x)) | a < x < f^{−1}(d)} If c ≤ f(a) < d, c < f(b) ≤ d then {(x, f(x)) | x ∈ R} ∩ (a, b) × [c, d) = ((a, f(a)), (b, f(b)) where ((a, f(a)), (b, f(b)) = {(x, f(x)) | a < x < b, } Thus, every open interval in L is open. Note that open intervals are open in the lower limit topology. Since we have determined for every possible intersection between L and basis elements of R × R_{l}, the collection of half open intervals is a basis for the topology on L. Thus L inherits the lower limit topology as a subspace of R × R_{l} Similarly, L inherits the lower limit topology as a subspace of R_{l} × R_{l} Similarly, if L is a line with negative slope, L inherits the upper limit topology as a subspace of R × R_{l} Let R^{2} have the topology generated by R_{l} × R_{l} Let L be a line with negative slope. L = {(x, y) | y = mx + b} where m < 0. Let f(x) = mx + b. [x, x + 1) × [f(x), f(x) + 1) ∩ L = {(x, f(x))}. Thus every point in L is open. Thus the topology on L is discrete topology.