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The Bolzano-Weierstrass theorem in $\mathbb{R}$ says that every bounded real sequence has a convergent subsequence. For a bounded sequence in ${\mathbb{R}}^n$, successively extract subsequences on which the first, second, and so on through the $n$th coordinate converge. The final diagonal subsequence converges in Euclidean space, proving the theorem in ${\mathbb{R}}^n$.

Put $K=D\setminus {\bigcup }_{z\in S}{B}_{\epsilon }(z)$. Suppose the asserted $\delta$ did not exist. Then there would be $x_k\in D$ and $y_k\in K$ with $\Vert x_k-y_k\Vert <1/k$ but $|f(x_k)-f(y_k)|\ge \epsilon$. Since $K$ is closed and bounded, Bolzano-Weierstrass gives a subsequence $y_k\to y\in K$, and then $x_k\to y$. The point $y$ is not in $S$, because otherwise $y\in B_{\epsilon }(y)$, so $f$ has continuity at $y$. Both $f(x_k)$ and $f(y_k)$ tend to $f(y)$, a contradiction. Hence the required uniform $\delta >0$ exists.

Choose a countable dense subset $\{x_1,x_2,\dots \}$ of the normed vector space $X$. For each fixed $j$, the scalar sequence $(f_n(x_j))$ is bounded becauseRepeated use of the Bolzano-Weierstrass theorem followed by a diagonal subsequence argument produces a subsequence $(f_n{)}_{n\in \Lambda }$ for which $f_n(x_j)$ converges for every $j$.

Let $D$ be the linear span of the $x_j$. On every $x\in D$, defineFinite linear combinations show that $f:D\to \mathbb{R}$ is linear, and passage to the limit in $|f_n(x)|\le \Vert x\Vert$ gives $|f(x)|\le \Vert x\Vert$. Thus $f$ is a bounded linear functional on the dense linear subspace $D$ and extends uniquely to an element of the dual space $X^{\ast }$ with $\Vert f\Vert \le 1$.

For arbitrary $x\in X$ and $y\in D$,First choose $y$ close to $x$, then let $n\in \Lambda$ tend to infinity. This provesThis is the Sequential Banach-Alaoglu theorem for a separable predual.

Condition (iii) is also sufficient. If $|a_n|$ did not tend to infinity, some $M>0$ would contain infinitely many terms with $|a_n|\le M$. Those terms form a bounded subsequence, and the Bolzano-Weierstrass theorem would give it a convergent subsequence, contrary to the hypothesis. Hence $|a_n|\to \infty$ and

The Bolzano-Weierstrass theorem states that every bounded sequence of real numbers has a convergent subsequence.

Condition (i) is insufficient because the allowed limit may be zero. For example, $a_n=1/n$ is nonzero and converges to zero, whereas $1/a_n=n$ does not converge in $\mathbb{R}$.

Choose a closed interval containing the bounded sequence. Bisect it and retain a half containing infinitely many terms; repeat. The resulting intervals are nested and their lengths tend to zero. By the nested interval theorem they have one common point $x$, and choosing successively indexed terms from these intervals produces a subsequence converging to $x$. This proves the Bolzano-Weierstrass theorem.

Boundedness of $(z_n)$ makes both real sequences $(x_n)$ and $(y_n)$ bounded. The Bolzano-Weierstrass theorem first gives a subsequence $(z_{n_j})$ for which $(x_{n_j})$ converges. Applying it again to $(y_{n_j})$ gives a subsubsequence for which the imaginary parts also converge. Since convergence of complex numbers is equivalent to convergence of their real parts and imaginary parts, this subsubsequence converges in $\mathbb{C}$.

Construct the nested index sequences inductively. Start with $N^{(0)}=(1,2,\dots )$. Having chosen $N^{(i-1)}$, apply the Bolzano-Weierstrass theorem to the bounded sequence of $i$th coordinates along those indices, and let $N^{(i)}$ be an increasing subsequence along which that coordinate converges. Because each new sequence is nested inside all earlier ones, coordinates $1,\dots ,i$ all converge along $N^{(i)}$.

Now take the diagonal subsequence argumentNestedness impliesso $(m_j)$ is strictly increasing. For every fixed $i$, the tail $(m_j{)}_{j\ge i}$ is a subsequence of $N^{(i)}$; hence $(x_{m_j}^{(i)})$ converges. This single diagonal subsequence works for every coordinate.

The Bolzano-Weierstrass theorem states that every bounded sequence of real numbers has a convergent subsequence.

For a proof, place all terms in a closed bounded interval $I_1$. Bisect it and choose a closed half $I_2$ containing infinitely many terms. Continue inductively, choosing nested closed intervalswith infinitely many sequence terms and with lengths tending to zero. Choose $n_k>n_{k-1}$ such that $x_{n_k}\in I_k$. The nested interval theorem gives a unique point $x$ in every $I_k$. Because both $x_{n_k}$ and $x$ lie in $I_k$,Thus $(x_{n_k})$ is a convergent subsequence.

Now suppose every convergent subsequence of the bounded sequence $(x_n)$ converges to $L$. If $x_n$ did not converge to $L$, there would be an $ϵ>0$ and a subsequence $(x_{n_k})$ satisfyingfor every $k$. This subsequence is bounded, so Bolzano--Weierstrass gives a convergent subsubsequence. By hypothesis its limit is $L$, contradicting the displayed inequality. Hence the unique subsequential limit of a bounded sequence principle gives

Let $\Vert x_n\Vert \le C$. For each $i$, Bessel inequality givesso the $i$th coordinate sequence is bounded. By the Bolzano-Weierstrass theorem, choose a subsequence on which the first coordinate converges. From it choose a further subsequence on which the second coordinate converges, and continue.

Take the diagonal subsequence $(x_{\varphi (n)})$. For every fixed $i$, its $i$th coordinate converges, and it remains bounded by $C$. The coordinate criterion for weak convergence in a separable Hilbert space supplies an $x_{\infty }\in H$ such thatThis proves the theorem on a weak subsequence of a bounded Hilbert-space sequence.

Let $f:[a,b]\to \mathbb{R}$ be continuous. If it were unbounded, one could choose $x_n\in [a,b]$ with $|f(x_n)|>n$. The Bolzano-Weierstrass theorem gives a subsequence $x_{n_j}\to x\in [a,b]$, but continuity would then give $f(x_{n_j})\to f(x)$, contradicting unboundedness.

Now let $M=\sup f([a,b])$. Choose $x_n$ with $f(x_n)>M-1/n$. A convergent subsequence and continuity give $f(x)=M$ at its limit. Applying the same argument to $-f$ shows that the infimum is attained. This proves the extreme value theorem on a closed bounded interval.

The functionis continuous and bounded, but attains neither its infimum $0$ nor its supremum $1$. For the second example, enumerate the rationals in $[0,1]$ as $(q_n)$ and setEvery nondegenerate interval contains infinitely many rational numbers, hence some $q_n$ with arbitrarily large $n$; therefore $\psi$ is unbounded on every such interval.

For the running extrema, compactness lets us writeThe function $f$ is uniformly continuous on $[a,b]$. Given $\epsilon >0$, choose $\delta >0$ such that $|f(u)-f(v)|<\epsilon$ whenever $|u-v|<\delta$. If $x<y$ and $y-x<\delta$, every new value $f(\xi )$ with $x\le \xi \le y$ is at most $f(x)+\epsilon \le M(x)+\epsilon$. Since $M(y)\ge M(x)$,Interchanging $x,y$ handles the other order. Applying this argument to $-f$ proves continuity of $m$ as well.

Finally fix $T>0$ and put $h(x)=g(x+T)-g(x)$. For each positive integer $n$, there must be some $x_n>n$ with $|h(x_n)|<1/n$. Otherwise, for some $n$ the continuous function $h$ would satisfy $|h(x)|\ge 1/n$ on the interval $(n,\infty )$. By the intermediate value theorem, $h$ would have a constant sign there. The valueswould then increase or decrease by at least $1/n$ at every step, contradicting boundedness of $g$. Thus $x_n\to \infty$ and