Uniqueness of Limits for Sequences and Subsequences in Metric Spaces

February 17, 2026 · Luciano Muratore

Uniqueness of Limits for Sequences and Subsequences in Metric Spaces

Let $(X, d)$ be a metric space.

Let $(x_n)$ be a sequence in $X$ that converges to $x \in X$, and let $(x_{n_k})$ be a subsequence of $(x_n)$ that converges to $y \in X$.

We prove that

Proof

Let $\varepsilon > 0$ be arbitrary.
We aim to show that

Step 1 — Triangle Inequalities

By the triangle inequality,

Again applying the triangle inequality,

Therefore,

Step 2 — Control Each Term

Since $x_n \to x$, there exists $N_1 \in \mathbb{N}$ such that for all $n > N_1$,

Since $x_{n_k} \to y$, there exists $N_2 \in \mathbb{N}$ such that for all $k > N_2$,

Because $(x_n)$ converges, it is a Cauchy sequence. Therefore, there exists $N_3 \in \mathbb{N}$ such that for all $n, k > N_3$,

Step 3 — Combine the Estimates

Let

Then for all $n, k > N$, we obtain

Step 4 — Conclude

Since $\varepsilon > 0$ was arbitrary, it follows that

Hence,

Why This Matters

This result formalizes an intuitive idea:

A subsequence cannot “escape” the limit of a convergent sequence.

It guarantees the uniqueness of limits in metric spaces and is a fundamental tool in:

• Real Analysis
• Topology
• Functional Analysis
• Compactness arguments

Without uniqueness, convergence theory would collapse.