Unformatted text preview: Brief solutions to the midterm.
1. The function d is a metric, since it satisﬁes all the metric axioms. (a) It is nonnegative for
any points p and q and is zero only if p = q. (b) The function is symmetric. (c) It satisﬁes
the triangle inequality. Indeed, if p = q, then d(p, q) = 0 does not exceed 0, 1 or 2, which
are possible on the righthand side. If p = q, then d(p, q) = 1 and the righthand side is at
least 1, since we cannot have p = r and q = r without having p = q.
The sets compact in this metric space are ﬁnite subsets F of X. They are obviously
compact, since any open cover of such a set contains a ﬁnite subcover: for each element
x ∈ F , just take the open set in the cover that contains x. These are the only compact sets,
since if a set S ⊂ X is inﬁnite, then a possible open cover for S is ∪x∈S {x}, which contains
no ﬁnite subcover.
2. The ﬁrst set is not open, closed, perfect, not connected, bounded. The second set is open,
not closed, not perfect, not connected, bounded.
3. The ﬁrst series diverges by comparison with the series n 1/n. The second converges,
being an alternating series (starting when log n > e). The terms of the third series can be
√
√
√
rewritten as 1/( 3 (n + 1)2 + 3 n + 1 · 3 n + 3 n2 ) and compared to 1/n2/3 . Since the series
2/3
diverges, so does the given series.
n 1/n
4. Let ε > 0 be given. Since (pn ) is a Cauchy sequence, there exists a number M = M (ε)
such that
d(pn , pm ) < ε/2 whenever n, m > M.
Since the subsequence (pnk ) converges to p, there exists K = K(ε) such that
p(pnk , p) < ε/2 whenever k > K. Let N := max{nK , M }. Then, for any n > N , we have
d(pn , p) ≤ d(pn , pnK+1 ) + d(pnK+1 , p) < ε/2 + ε/2 = ε.
So, limn→∞ pn = p. 1 ...
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This document was uploaded on 11/17/2010.
 Spring '09

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