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27) Let (X,d) be a metric space and (a n ​ ) n=1 [infinity] ​ be a sequence of elements of X. MAKE precise the following statement : a n ​ converges to L∈X if nd only if every subsequence of (a n ​ ) n=1 [infinity] ​ has a sub-sub-sequence which converges to L.

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Answer:

The statement "aₙ converges to L ∈ X if and only if every subsequence of (aₙ)ₙ₌₁^∞ has a sub-subsequence which converges to L" can be formalized as follows:

Definition 1: Convergence of a sequence

A sequence (aₙ)ₙ₌₁^∞ in a metric space (X, d) is said to converge to a point L ∈ X if for every positive real number ɛ > 0, there exists a positive integer N such that for all n ≥ N, the distance between aₙ and L, denoted as d(aₙ, L), is less than ɛ.

Definition 2: Subsequence

A subsequence of a sequence (aₙ)ₙ₌₁^∞ is obtained by selecting an infinite number of terms from the original sequence in their given order.

Definition 3: Sub-subsequence

A sub-subsequence of a subsequence (bₙₖ)ₖ₌₁^∞ is obtained by selecting an infinite number of terms from the subsequence in their given order.

Definition 4: Convergence of a sub-subsequence

A sub-subsequence (bₙₖₗ)ₗ₌₁^∞ of a subsequence (bₙₖ)ₖ₌₁^∞ is said to converge to a point L ∈ X if for every positive real number ɛ > 0, there exists a positive integer N such that for all l ≥ N, the distance between bₙₖₗ and L, denoted as d(bₙₖₗ, L), is less than ɛ.

The statement can now be formally stated as:

Statement: A sequence (aₙ)ₙ₌₁^∞ in a metric space (X, d) converges to a point L ∈ X if and only if every subsequence of (aₙ)ₙ₌₁^∞ has a sub-subsequence that converges to L.

In simpler terms, this statement says that a sequence converges to a certain point if and only if every subsequence of that sequence contains a further subsequence that converges to the same point.

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User Simon Hobbs
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