(2) ℝω in the uniform topology is disconnected. Proof: The sets of bounded and unbounded sequences in ℝω are both open in this metric, and form a separation.
(3) The ordered square is not path connected. Proof: Suppose there exists a path from 0×0 to 1×1. Since I20 is a linear continuum, this implies the path is surjective and I20 is an image of the separable [0,1]. But I20 itself is not separable, for r×(1/4,3/4) for r∈I is a collection of uncountably many disjoint open subsets.
(4) ℝK is not path connected. Proof: Suppose f:[0,1]→ℝK is a path from 0 to 1. Then f−1(0) is a closed set not containing 1, so r=sup f−1(0)<1. Since [r,1]≅[0,1] we may assume f(x)>0 for x>0.
Now, let an=inf f−1(1/n) for n∈ℕ+. We see an+1<an by connectivity of continuous images, but also an↛0 in [0,1] since 1/n↛0 in ℝK. Hence let a>0 be such that a<an for all n. But then f(a)>0 so 1/N<f(a) for some N, implying aN<a again by connectivity of continuous images, a contradiction.
(5) The ordered square is not locally path connected. Proof: The proof of (3) extends to show that the path components of I20 are precisely r×[0,1] for r∈[0,1], which are not open.
(6) ℝl is not locally compact at any of its points. Proof: Let x∈ℝl and U be a hood of x. Suppose V is a hood of x such that ¯V⊆U is compact. Then x∈[a,b)⊆V for some a,b∈ℝ, and since [a,b) is also closed in ¯V this implies [a,b) is compact. However, [a,b) is not compact even in ℝ.
(7) ℝω in the uniform topology is not locally compact. Proof: Suppose C is a compact subset of ℝω containing a hood of i=(0,0,...). Then B[i,ε]=[−ε,ε]ω is compact for some ε∈(0,1). But {en}n∈ℕ+ (when en is the point with zeros in every coordinate except the nth in which it is ε) is an infinite subset containing no limit point (as d(en,em)=ε for every n≠m), so [−ε,ε]ω cannot even be limit point compact.
(8) ℝω in the uniform topology is not second countable, separable, or Lindelof. Proof: We see each pair of distinct x,y∈{0,1}ω are of distance 1 in ℝω, so that ℝω cannot be separable and hence not second countable. As well, {B(x,3/4) | x∈{0,1}ω} is an uncountable open cover of the closed subset [0,1]ω, yet there are not even any proper subcovers, so ℝω cannot be Lindelof.
(9) ℝI is not locally metrizable. Proof: Suppose some basis element U=∏i∈IUi of ℝI were metrizable. Then since Ui=ℝ for all but finitely many i∈I, and since I−F for any finite subset F⊆I is still uncountably infinite, we see ℝI can be imbedded in U. But ℝI itself is not metrizable, as it is not normal.
(10) ℝI is not Lindelof. Proof: Regular Lindelof spaces are normal, which ℝI is not.
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