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Every bounded sequence of real numbers contains a subsequence with a finite limit of a sequence. If $|x_n|\le M$ for every positive integer $n$, there are integers $n_1<n_2<\cdots$ and a real number $\ell$ such thatThe increasing indices matter: a subsequence retains the original order and selects terms arbitrarily far along the original sequence. The theorem guarantees the existence of at least one such selection; it does not claim that the entire sequence converges or that the subsequential limit is unique.^[ref]

For example, the bounded sequence $x_n=(-1{)}^n+1/(n+1)$ has two distinct subsequential limits:The even terms form a convergent sequence, and so do the odd terms. The full sequence cannot have a limit of a sequence because its two subsequences have different limits. Its limit superior is $1$ and its limit inferior is $-1$.

This is useful when a construction produces many candidate points but does not immediately produce a convergent sequence. For a continuous function on a nonempty closed bounded feasible region in a finite-dimensional Euclidean space, take a sequence whose objective values approach the infimum. A convergent subsequence supplies a candidate point; a closed subset retains that point, and continuity passes the objective values to the limit of a sequence. This is the mechanism behind attainment in the extreme value theorem. The diagonal subsequence argument extends the real-sequence theorem to finite-dimensional Euclidean space.

The theorem is named for Bernard Bolzano and Karl Weierstrass.^[ref]

The limit inferior and limit superior retain the lower and upper long-run behaviour of a real sequence, even when it has no ordinary limit of a sequence. For each tail of a sequence beginning at $N$, defineDiscarding more initial terms raises or preserves the infimum and lowers or preserves the supremum. Hence $(l_N)$ is nondecreasing and $(u_N)$ is nonincreasing. Their limits exist in the extended real number line:For a bounded sequence, both values are finite and $\mathrm{liminf}{a}_n\le \mathrm{limsup}{a}_n$. These bounds use the entire infinite tail of a sequence, rather than the last few observed terms. ^[ref].

Take the original example $a_n=(-1{)}^n(1+1/n)$, beginning $-2,3/2,-4/3,5/4,\dots$. The subsequence at even integers decreases toward one, while the subsequence at odd integers increases toward minus one. If $e_N$ is the first even integer at least $N$ and $o_N$ is the first odd integer at least $N$, thenThus $\mathrm{liminf}{a}_n=-1$ and $\mathrm{limsup}{a}_n=1$. The finite initial extreme $a_1=-2$ disappears from later lower bounds; neither long-run value is attained by any term of this example.

For a bounded sequence, the limit inferior is the smallest subsequential limit and the limit superior is the largest subsequential limit; both are reached by suitable convergent subsequences. Every other subsequential limit lies between them. In the example there are only two subsequential limits, $-1$ and $1$: a convergent subsequence cannot retain infinitely many terms of both parities, because those two branches stay separated. The whole interval between the two bounds need not consist of subsequential limits. ^[ref].

A finite ordinary limit of a sequence exists exactly when the limit inferior and limit superior agree at the same finite value $L$. The bounds $l_N\le a_N\le u_N$ and the squeeze theorem prove the sufficient direction. For a bounded sequence, its tail diameter of a sequence is $D_N=u_N-l_N$, so the limiting gap is zero exactly when the Cauchy sequence condition holds. In the example the gap approaches two, not zero. The same framework allows infinite bounds: $a_n=n$ has both values $+\infty$, whereas $a_n=(-1{)}^{n}n$ has limit inferior $-\infty$ and limit superior $+\infty$. Equality at infinity describes escape rather than convergence to a real value. ^[ref].