# 6 - Math 310 hw 6 solutions Wednesday 21 Oct 2009 11.6 11.7...

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Math 310 - hw 6 solutions 11.6, 11.7, 11.10; Wednesday, 21 Oct 2009 11.6 Show that every subsequence of a subsequence of a given sequence is itself a subsequence of the given sequence Hint: Deﬁne subsequence as in (3) of Deﬁnition 11.1. Let ( s n ) n be a sequence which we regard as a function s : N R . A subsequence ( s n k ) k may be regarded as in (3) of Deﬁnition 11.1, as a composition s σ : N R where σ : N N , given for k N , by σ ( k ) = n k , is increasing, i.e. σ ( k ) < σ ( k + 1). A subsequence of this subsequence is then given by ( s σ ) τ where τ : N N is another increasing function. We need to show that σ τ : N N is also an increasing function. Let m N . We ﬁrst prove by induction that for all n N , τ ( m ) < τ ( m + n ). Since σ is increasing, σ ( m ) < σ ( m + 1), so the base case holds. Next n N and suppose that σ ( m ) < σ ( m + n ). Since σ is increasing, σ ( m + n ) < σ ( m + n + 1). Thus by transitivity
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## This note was uploaded on 03/26/2010 for the course MATHEMARIC 131a taught by Professor Janeday during the Spring '10 term at San Jose State.

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