Bolzano Weierstrass Theorem¶

Theorem¶

Each bounded sequence in the Euclidean space $(\mathbb{R}^n,d_2)$ has a convergent subsequence.

Equivalently, a subset $E$ of the Euclidean space is sequentially compact if and only if it is both closed and bounded.

Interpretation¶

Even if a sequence wanders around without settling down, as long as it stays within a bounded region you can always pick out a subsequence that homes in on some point.

All terms lie in the band $|a_n|\le M$$M$, yet the sequence does not converge: its terms keep jumping between the two halves of the band. The even-indexed terms, in red, form a subsequence that converges to $1$, while the odd-indexed terms stay below the axis. The theorem promises at least one such subsequence; it does not say where its limit lies.

In $\mathbb{R}$ the reason is repeated halving. One of the two halves of $[-M,M]$ contains terms $a_n$ for infinitely many $n$, one half of that half again, and so on; the nested intervals shrink to a point, and one term from each interval gives a subsequence converging to it. In $\mathbb{R}^n$ the argument runs coordinate by coordinate.

Both parts of the setting are needed. Without boundedness the conclusion fails: $x_n=n$ has no convergent subsequence. Without the Euclidean metric it can fail too: under the discrete metric every set is bounded, but a sequence of distinct points has no convergent subsequence.

The limits of convergent subsequences are the accumulation points of the sequence, so a bounded sequence in $\mathbb{R}^n$ has at least one accumulation point. In the second form, closedness keeps the limit inside $E$, so closed and bounded sets are sequentially compact; with the equivalence of compactness and sequential compactness in metric spaces, this gives the Heine-Borel theorem.

Key Results¶

• The limit of a convergent subsequence is an accumulation point of the original sequence.
• In $\mathbb{R}^n$, every Cauchy sequence is bounded, so by Bolzano–Weierstrass theorem it has a convergent subsequence. A Cauchy sequence with a convergent subsequence must itself converge. This gives another proof that $\mathbb{R}^n$ is complete.

Examples¶

• The sequence $x_n = (-1)^n$ in $\mathbb{R}$ is bounded between $-1$ and $1$. It does not converge, but it has two convergent subsequences: the even terms $x_{2n}=1$ converge to $1$, and the odd terms $x_{2n-1}=-1$ converge to $-1$.
• The sequence $x_n = 1/n$ in $\mathbb{R}$ is bounded in $[0,1]$. It already converges to $0$, so every subsequence also converges to $0$.

Prerequisites¶

• Sequentially Compact Set
• Open and Closed Sets in a Metric Space
• Bounded Set in a Metric Space

Related Concepts¶

• Heine-Borel Theorem
• Subsequence

Sources¶

• Math III: Analysis · Lecture 9

Aliases: Bolzano-Weierstrass theorem Keywords: sequential compactness theorem Symbols: real numbers

Last updated: 20 October 2026 07:29