1st December 2004

Munkres §16

Ex. 16.1 (Morten Poulsen). Let (X, T ) be a topological space, (Y, T_{Y}) be a subspace and let A ⊂ Y.

Let T_{A}^{Y} be the subspace topology on A as a subset of Y and let T_{A}^{X} be the subspace topology on A as a subset of X. Since

U ∈ T_{A}^{Y} ⇔ ∃ U_{Y} ∈ T_{Y} : U = A ∩ U_{Y}
⇔ ∃ U_{X} ∈ T : U = A ∩ (Y ∩ U_{X})
⇔ ∃ U_{X} ∈ T : U = A ∩ U_{X}
⇔ U ∈ T_{A}^{X}

it follows that T_{A}^{Y} = T_{A}^{X}.

Ex. 16.3 (Morten Poulsen). Consider Y = [−1, 1] as a subspace of R with the standard topology. By lemma 16.1 a basis for the subspace topology on Y is sets of the form:

Y ∩ (a, b) = \{ \begin{matrix}(a, b), & a, b ∈ Y \\ [−1, b), & a ∉ Y, b ∈ Y \\ (a, 1], & a ∈ Y, b ∉ Y \\ Y, ∅, & a, b ∉ Y. \end{matrix}

Note that intervals of the form [a, b) are not open in R, since there are no basis element (c, d) such that a ∈ (c, d) ⊂ [a, b). Similarly are intervals of the form (a, b] and [a, b] not open in R.

A = \{ x | \frac{1}{2} < |x| < 1 \}: A = (−1, −\frac{1}{2}) ∪ (\frac{1}{2}, 1), hence A open in R. Since A = Y ∩ A it follows that A open in Y.

B = \{ x | \frac{1}{2} < |x| ≤ 1 \}: Since B = Y ∩ (−2, −\frac{1}{2}) ∪ (\frac{1}{2}, 2) it follows that B open in Y. Another argument is that B = [−1, −\frac{1}{2}) ∪ (\frac{1}{2}, 1], i.e. an union of basis elements, hence open in Y. The set B is not open in R, since if B is open in R then B ∩ (0, 2) = (\frac{1}{2}, 1] open in R, contradicting that (\frac{1}{2}, 1] not open in R.

C = \{ x | \frac{1}{2} ≤ |x| < 1 \}: Since there clearly is no basis element U for the subspace topology on Y such that \frac{1}{2} ∈ U ⊂ C, it follows that C is not open in Y. By an argument similar to the one for the set B, it follows that C not open in R.

D = \{ x | \frac{1}{2} ≤ |x| ≤ 1 \}: By arguments similar to the ones above it is easily seen that D is not open in either Y or R.

E = \{ x | 0 < |x| < 1, \frac{1}{x} ∉ Z_{+} \}: Note that E = (−1, 0) ∪ ((0, 1) − K), where K = \{\frac{1}{n} | n ∈ Z_{+} \}. Since E = (−1, 0) ∪ ⋃_{n=1}^{∞}(\frac{1}{n+1}, \frac{1}{n}) it follows that E is open in R and Y.

Ex. 16.4. Let π_{k}: ∏_{j∈J} X_{j} → X_{k} be the projection map onto X_{k}. Observe that π_{k} maps any basis set, ∏ V_{j} where V_{j} ⊂ X_{j} is open and V_{j} = X_{j} for all but finitely many j ∈ J, to an open set in X_{k}, π_{k}(∏ V_{j}) = V_{k}. Since maps preserve unions of sets [Ex 2.2], it follows that π_{k} maps open sets in the product ∏ X_{j} to open sets in X_{k}.

Ex. 16.6 (Morten Poulsen). The set

B = { (a, b) × (c, d) | a < b, c < d and a, b, c, d ∈ Q }

is a basis for R^{2}: The set

{ (r, s) | r < s and r, s ∈ Q }

is a basis for R, by Ex. 13.8(a). From Theorem 15.1 it follows that B is a basis for R^{2}.

Ex. 16.7. R_{+} × R is a convex subset of the linearly ordered set R × R that is not an interval nor a ray.

Ex. 16.9 (Morten Poulsen). Let R_{dict}^{2} be R^{2} with the dictionary order topology and let R_{d} × R be the product topology, where R_{d} is R with the discrete topology and R is R with the standard topology.

The set

{ (a × b, c × d) | a, b ∈ R and (a < c) ∨ (a = c ∧ b < d) }

is a basis for R_{dict}^{2}.

The set

{ {a} | a ∈ R }

is a basis for R_{d}, c.f. §13 Example 3.

The set

{ (a, b) | a, b ∈ R and a < b }

is a basis for R.

By Theorem 15.1 the set

{ {a} × (b, c) | a, b, c ∈ R and b < c }

is a basis for R_{d} × R.

Claim 1. R_{dict}^{2} = R_{d} × R.

Proof. ”⊂”: Given basis element (a × b, c × d) ∈ R_{dict}^{2} and x × y ∈ (a × b, c × d) then x × y ∈ {x} × I ⊂ (a × b, c × d), where I is an open interval in R containing y. By Lemma 13.3 it follows that R_{d} × R is finer than R_{dict}^{2}.

”⊃”: Clear, since every basis element {a} × (b, c) ∈ R_{d} × R is a basis element in R_{dict}^{2}. □

References