Eksamen januar 2005

JMM

Solutions to the January 2005 exam

Problem 1

(1) ⟹ (2): Since f is continuous f^{−1}(B^{∘}) ⊂ f^{−1}(B)^{∘} [2, 18.1] [1, 13.3]. The other inclusion, f^{−1}(B)^{∘} ⊂ f^{−1}(B^{∘}), equivalent to f(f^{−1}(B)^{∘}) ⊂ B^{∘}, follows because f(f^{−1}(B)^{∘}) is open and contained in f(f^{−1}(B))) ⊂ B.

(2) ⟹ (3): X − f^{−1}(\overline{B}) = f^{−1}(Y − \overline{B}) = f^{−1}((Y − B)^{∘}) =^{(2)} f^{−1}(Y − B)^{∘} = (X − f^{−1}(B))^{∘} =

X − \overline{f^{−1}(B)} so that f^{−1}(\overline{B}) = \overline{f^{−1}(B)} for all B ⊂ Y.

(3) ⟹ (2): X − f^{−1}(B)^{∘} = \overline{X − f^{−1}(B)} = \overline{f^{−1}(Y − B)} =^{(3)} f^{−1}(\overline{Y − B}) = f^{−1}(Y − B^{∘}) =

X − f^{−1}(B^{∘}) so that f^{−1}(B^{∘}) = f^{−1}(B)^{∘} for all B ⊂ Y.

(2) ⟹ (1): The inclusion f^{−1}(B^{∘}) ⊂ f^{−1}(B)^{∘}, valid for all B ⊂ Y, tells us that f is continuous.

Since A^{∘} ⊂ f^{−1}f(A)^{∘} =^{(2)} f^{−1}(f(A)^{∘}), we have f(A^{∘}) ⊂ f(A)^{∘} for all A ⊂ X. In particular, f(A) = f(A^{∘}) ⊂ f(A)^{∘} ⊂ f(A), ie f(A) = f(A)^{∘}, when A is open. Thus f is open. Problem 2

(1) Ext(A) is open because it is the complement of the closed set \overline{A}. Bd A is closed because it is the intersection of two closed sets.

(2) X − (Int(A) ∪ Ext(A)) = (X − Int(A)) ∩ \overline{A} = \overline{X − A} ∩ \overline{A} = Bd(A) where Int(A) ∩ Ext(A) ⊂ A ∩ (X − \overline{A}) ⊂ A ∩ (X − A) = ∅.

(3) Assume that u(I) ∩ Bd(A) = ∅. Then [0, 1] = u^{−1}(Int A) ∪ u^{−1}(Ext A) is a union of two disjoint open nonempty sets, contradicting connectedness [2, §23] [1, 16.2] of the unit interval [2, 24.2] [1, 17.3].

Problem 3

(1) Consider the straight line L_{n} ⊂ I ×I between the points \frac{1}{n+1} ×1 and \frac{1}{n} ×1 of I ×\{0\}∪A×I. The image R(L_{n}) is a connected subspace [1, 16.3] of I × {0} ∪ A × I containing the two end-points of L_{n}. Since I × \{0\} ∪ A × I with the point P_{n} = \frac{1}{2} (\frac{1}{n+1} + \frac{1}{n}) × 0 removed is not connected, R(L_{n}) contains P_{n}.

(2) The sequence t_{n} × 1 converges to 0 × 1 and the image sequence R(t_{n} × 1) = P_{n} converges to 0 × 0 ≠ 0 × 1 = R(0 × 1) contradicting continuity of R [1, 15.12].

Alternative solution: Assume that the map R exists. Then R is a quotient map because it is a map from a compact space onto a Hausdorff space [1, 18.8]. But I × {0} ∪ A × I is not locally connected [1, 17.17] so it can not be the quotient space of the locally connected space I × I. Problem 4

(1) T ∩X_{n} is closed for all n because it is a finite set in a Hausdorff space. Since the topology on X is coherent with the filtration, T is closed. The same argument applies to any subspace of T. Since any subspace of T is closed, T has the discrete topology.

(2) Closed, discrete subspaces of compact spaces are finite [1, 18.13].

(3) Since T is finite there is an N such that C ∩ (X_{n+1} − X_{n}) = ∅ for all n ≥ N. This means that C ⊂ X_{N}.

References

[1] Jesper M. Møller, General topology, http://www.math.ku.dk/ moller/e03/3gt/notes/gtnotes.dvi.

[2] James R. Munkres, Topology. Second edition, Prentice-Hall Inc., Englewood Cliffs, N.J., 2000. MR 57 #4063