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Homework_4_Solutions2

Homework_4_Solutions2 - ⊂ B then A open = ⇒ A B open...

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3.2.3.15 Let U R be an open set. For every x U , let I x = ( a, b ), where a = inf { y < x : [ y, x ] U } and b = sup { y > x : [ x, y ] U } . Both sets above are non-empty since U is open; if { y < x : [ y, x ] U } isn’t bounded below, set a = -∞ , and if { y > x : [ x, y ] U } isn’t bounded above, set b = . I claim that all I x are intervals, open by definition, and that for all x, y U we have either I x = I y or I x I y = . If y, z I x with y < z , and y c z , then either c = x in which case c I x , or y c < x , so [ c, x ] [ y, x ] ( a, x ] I x , or x < c z so [ x, c ] [ x, z ] [ x, b ) I x . Suppose that z I x I y . Set I x = ( a, b ) and I y = ( c, d ). Without loss of generality, we may suppose that a c z Then all the numbers u ( a, c ] are in U , since they are in ( a, z ] I x . If a < c this contradicts the definition of I y . Thus a = c . The proof that b = d is similar. 3.3.1.7 If A is open and B is any set, then A · B may not be open. This happens if 0 B but no neighborhood of 0 is in B . For instance (1 , 2) · [0 , 1) = [0 , 2). But if either 0 / B or there exists > 0 such that (
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Unformatted text preview: ) ⊂ B , then A open = ⇒ A · B open . Let a ∈ A and b ∈ B . If b 6 = 0, and ( a-±, a + ± ) ⊂ A for some ± > 0, then ( ab- | b | ±, ab + | b | ± ) ⊂ A · B. Now suppose b = 0 and (-±, ± ) ⊂ B for some ± > 0. If a ∈ A satisfies a 6 = 0, then (-| a | ±, | a | ± ) ⊂ A · B , providing a neighborhood of ab = 0 contained in A · B in that case. If a = 0, then there exists δ > 0 with (-δ, δ ) ⊂ A , and then (-±δ, ±δ ) ⊂ A · B . 4.1.5.2 The set A is closed because A = A 1 ∩ A 2 ∩ ··· ∩ A n , where A i is the set defined by f i ( x ) = 0, and each A i is the inverse image of the closed set { } by a continuous map, hence closed. It need not be compact, of course: consider the case n = 1 and f ( x ) = sin πx , so A = Z . 1...
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