Limit of sequence in complete space well-defined?

Question

Given a metric space $X$, two complete metric spaces $X'$ and $Y$, and isometries $i : X \rightarrow X'$ and $f : X \rightarrow Y$, suppose I have a sequence $(i(x_n))_n$ in $i[X]$ that converges to some $p ∈ \overline{i[X]} = X'$. Then this sequence is Cauchy, and therefore $(x_n)_n$ is also Cauchy (in $X$). Hence $(f(x_n))_n$ is also Cauchy (in $Y$), and as $Y$ is complete, it converges to some $q ∈ Y$. What I want to know is: if $(i(y_n))_n$ is another sequence that converges to $p$, does that guarantee that $(f(y_n))_n$ converges to the same limit $q$?

Answer 1

Yes. Consider the sequence $z_n$ in $X$ defined by $z_{2n} = x_n$ and $z_{2n+1} = y_n$.

First you can check that $i(z_n) \to p$ in $X'$. To do this, for a fixed $\varepsilon > 0$ we need to find $N$ such that $n \geq N$ implies that $d(i(z_n), p) < \varepsilon$ where the distance is in $X'$. Since $i(z_{2n}) \to p$ and $i(z_{2n+1}) \to p$ there are numbers $N_0$ and $N_1$ such that $n \geq N_0$ implies that $d(i(z_{2n}),p) < \varepsilon$ and $n \geq N_1$ implies that $d(i(z_{2n+1}),p)< \varepsilon$. This means that we can choose $N = \max\{2N_0, 2N_1+1\}$.

From this convergence, you can conclude that $z_n$ is Cauchy in $X$ in the same way that you concluded that $x_n$ was.

This then means that $f(z_n)$ is Cauchy in $Y$ and hence converges. Suppose that $f(z_n) \to q' \in Y$. Since subsequences of a convergent sequence converge to the same limit we then have that $f(z_{2n}) \to q'$. By the definition of $z_{2n}$ and of $q$, we have that $f(z_{2n}) \to q$ also. Hence by the uniqueness of limits in a metric space, we can conclude that $q = q'$. Putting this together, we have that $f(z_n) \to q$ and in particular that $f(y_n) = f(z_{2n+1}) \to q$ also.

Answer 2

Yes; this follows most easily from a lemma valid in any metric space $(M,d)$:

If $a_n$ and $b_n$ are two sequences such that $\lim_{n\rightarrow\infty} d(a_n,b_n) = 0$ then the limits $\lim_{n\rightarrow\infty}a_n$ and $\lim_{n\rightarrow\infty}b_n$ are either equal or both non-existent.

To prove this, it suffices to assume that $a_n$ converges to some limit $A$ and then to prove that $b_n$ converges to $A$. However, this is easy since we can choose some $N$ such that if $n>N$ then it is true both that $d(a_n,A)<\varepsilon/2$ and $d(a_n,b)<\varepsilon/2$ by the hypothesis and henvce that $d(b_n,A) < \varepsilon$ by the triangle inequality, so $b_n$ converges to $A$.

The (partial) converse to this is also true that if $a_n$ and $b_n$ share a limit, then $d(a_n,b_n)=0$. It's also worth noting that this condition of decreasing distance defines an equivalence relation on the set of sequences - and the "standard" completion is often defined as the set of Cauchy sequences modulo this relation.

This lemma makes your problem easy, since it encodes everything you want to reason about in a statement about $X$. In particular, since $i(x_n)$ and $i(y_n)$ converge to the same thing, it must be that the distance $d(i(x_n),i(y_n))=d(x_n,y_n)=d(f(x_n),f(y_n))$ tends to zero, immediately giving you the result without any difficult analysis.