Question

Example of topological space in which sequence (n) comverges to every real number

Answer 1

Answer:

Step-by-step explanation:

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Let X be a topological space, ⟨xn:n∈N⟩ a sequence of points of X, and x∈X.

Definition 1. ⟨xn:n∈N⟩ converges to x if and only if for each open nbhd U of x there is an m∈N such that xn∈U whenever n≥mU.

We sometimes express this informally by saying that every open nbhd of x contains a tail of the sequence, meaning all of the terms from some point on.

This is a generalization of the familiar definition of convergence of sequences of real numbers:

Definition 2. ⟨xn:n∈N⟩→x if and only if for each ϵ>0 there is an mϵ∈N such that xn∈(x−ϵ,x+ϵ) whenever n≥mϵ.

The sets (x−ϵ,x+ϵ) for positive ϵ aren’t the only open nbhds of x in the usual topology on R, but it turns out that if U is an open nbhd of x in the usual topology on R, then there is an ϵ>0 such that (x−ϵ,x+ϵ)⊆U, and we can use this fact to show that a sequence of real numbers converges by one definition if and only if it converges by the other. Thus, you know lots of examples of non-convergent sequences in R with its usual topology; xn=(−1)n and xn=n give you two of them.

Here’s another fairly simple example. Let X=R, and let

τ={U⊆X:X∖U is a countable set}∪{∅}.

Check that ⟨X,τ⟩ is a topological space. (This topology τ is called the co-countable topology on X.) Let ⟨xn:n∈N⟩ be any sequence of distinct points of X, and let x be any point of X; I claim that ⟨xn:n∈N⟩ does not converge to x. To see this, let S={xn:n∈N}∖{x}, and let U=X∖S. Then X∖U=S, which is countable, so U∈τ, i.e., U is an open set in this space. Moreover, x∉S, so x∈U, and U is therefore an open nbhd of x. Clearly no xn belongs to the set U, so there cannot possibly be an mU∈N such that xn∈U whenever n≥mU, and therefore ⟨xn:n∈N⟩ does not converge to x. (In fact it turns out that this is another space in which the only convergent sequences are the ones that are constant from some point on; you might try to prove that.)

In your specific problem the topology of X is the discrete topology, meaning that every subset of X is an open set, and you’re given that the sequence ⟨an:n∈N⟩ converges to b in X. Since every subset of X is open, {b} is an open nbhd of b; and since ⟨an:n∈N⟩ converges to b, there must be some m∈N such that an∈{b} whenever n≥m. If an∈{b}, what is an