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A matrix summability method is regular if it preserves the ordinary limit of every convergent sequence. The Silverman-Toeplitz theorem characterizes regularity by uniformly bounded absolute row sums, entrywise convergence to zero down every fixed column, and convergence of the row sums to one.

Let $\mathbb{N}=\{1,2,\dots \}$ and give each factor of ${\mathbb{R}}^{\mathbb{N}}$ the usual topology of the real line. The familiar standard basis vectors extend to unit sequences $e_n$ with $(e_n{)}_j=1$ when $j=n$ and zero otherwise. Their finite linear combinations are exactly the finitely supported sequences, rather than all of ${\mathbb{R}}^{\mathbb{N}}$.

For every fixed coordinate $j$, $(e_n{)}_j=0$ for all $n>j$. Therefore $e_n\to 0$ in the product topology. More directly, any basic cylinder set $B(F,U)$ containing zero contains every $e_n$ with $n>\max (F\cup \{0\})$: all of its tested coordinates are then zero. The moving unit entry escapes every fixed finite collection of observations.

In the l-infinity sequence space of bounded real sequences, however,Thus the same sequence does not converge to zero in the supremum norm. Here the comparison is between the subspace topology inherited from the product topology and the norm topology on ${\ell }^{\infty }$. The supremum norm is not a finite-valued norm on all of ${\mathbb{R}}^{\mathbb{N}}$, which also contains unbounded sequences. In the box topology on ${\mathbb{R}}^{\mathbb{N}}$, the neighborhood ${\prod }_{j\ge 1}(-1/2,1/2)$ of zero excludes every $e_n$, so $e_n$ is not a convergent sequence with limit zero there either.

For a function with derivatives of every order at $a$, its Taylor series is the power seriesIts partial sums are Taylor polynomials. At $a=0$ it is called a Maclaurin series. Forming the coefficients, proving that its partial sums form a convergent sequence, and proving that their limit equals $f(x)$ are separate tasks. A smooth function provides all the coefficients, but a real analytic function is additionally represented locally by the resulting power series.

For the exponential function, every derivative of $e^x$ at zero equals one, givingHere equality holds for every real $x$: the Taylor remainder is bounded by $e^{|x|}|x{|}^{N+1}/(N+1)!$, which tends to zero for each fixed $x$. At $x=1$, the degree-five Taylor polynomial is $163/60\approx 2.71666667$, and the absolute difference from $e$ is less than $e/720$. The factorial denominator eventually overwhelms every fixed power, allowing local derivative information to reconstruct this function throughout the real line.

The geometric series shows why the region of validity matters. For $f(x)=1/(1-x)$, the Taylor series at zero is ${\sum }_{k\ge 0}{x}^k$, with radius of convergence one. For $x\ne 1$, its finite Taylor polynomial satisfies the exact identityAt $x=1/2$ and $N=5$, the remainder is $1/32$. At $x=0.9$, the same degree-five polynomial has value $4.68559$ and remainder $5.31441$, although the Taylor series still converges to $10$ there. Convergence of the infinite expression does not guarantee accuracy from a particular small number of terms. For $|x|>1$, even the terms fail to tend to zero, so this power series cannot represent the function. The standard expansions and the conditions for convergent series are discussed in OpenStax, working with Taylor series.

A stronger caution comes from the flat functionIt is a smooth function with $h^{(n)}(0)=0$ for every $n$, so every Taylor polynomial at zero is zero and its Taylor series converges everywhere to zero. Nevertheless $h(1/2)=e^{-4}>0$. Off zero, repeated differentiation produces a polynomial in $1/x$ times $e^{-1/x^2}$; the exponential decay makes each required limit at zero vanish. This explains how all the local coefficients can miss a positive function arbitrarily close to the centre. Real analyticity supplies the additional condition that rules out this behavior.

A subset of the plane can have three successive strict changes under taking its set of accumulation points. For example letPut $B=\{(1/n+1/(n^{2}m),0):n,m\ge 1\}$ and $C=\{(1/n,0):n\ge 1\}$. Then $A^{\prime }=B\cup C\cup \{0\}$, $A^{\prime \prime }=C\cup \{0\}$ and $A^{\prime \prime \prime }=\{0\}$, where $0$ denotes the origin. For fixed $n,m$, increasing $k$ gives a point of $B$; increasing $m$ gives a point of $C$; increasing $n$ gives the origin. These are all possible limits: in any convergent sequence of distinct points, either $n$ is unbounded, or a subsequence fixes $n$ and has unbounded $m$, or a subsequence fixes both and has unbounded $k$. The first-coordinate cluster intervals $[1/n,1/n+1/n^2]$ are disjoint and separated away from the origin. Repeating this classification after each differentiation proves the displayed derived sets. Each derivative removes one layer of isolated points, and all three displayed changes are strict.

A uniformly continuous function has one input tolerance valid throughout its domain of a function. For $f:D\to \mathbb{R}$, with $D\subseteq \mathbb{R}$, uniform continuity meansChoose the output tolerance $\epsilon$ first, then a positive $\delta$ valid for every pair of function inputs. It may depend on $\epsilon$, the function and its domain of a function, but not on the pair’s location. The absolute values measure input and output distance.^[ref]

Compare the quantifier order with the epsilon-delta definition of continuity:Both lines end with $|x-a|<\delta \Rightarrow |f(x)-f(a)|<\epsilon$. In the first, $\delta$ can depend on the point $a$; in the second, it is chosen before $a$. Thus uniform continuity implies continuity at every point. Each requested $\epsilon$ gets its own shared tolerance.

For an exact example, take the polynomial function $f(x)=x^2$ on the closed interval $D=[0,2]$. For every pair of function inputs in that interval, polynomial factorization givesConsequently $\delta =\epsilon /4$ works everywhere on this domain of a function. With $\epsilon =0.4$, the common tolerance is $\delta =0.1$. Near the right endpoint, for example, $x=1.92$ and $y=2$ have input distance $0.08$ and output distance $4-1.92^2=0.3136<0.4$. The bound is sufficient; it does not claim the largest admissible input window at each point.

On the whole real line, the same polynomial function is a continuous function but fails uniform continuity. Take the sequences $x_n=n$ and $y_n=n+1/n$, for positive integers $n$. ThenFor any proposed $\delta >0$, choose an integer $n>1/\delta$. The inputs are closer than $\delta$, yet their outputs violate the fixed tolerance $\epsilon =1$. The sequences need not themselves have a finite limit: their mutual distance is what tends to zero. This is the obstruction captured by the sequential criterion for uniform continuity. Restricting the domain of a function to $[0,2]$ removes these escaping pairs.

A uniformly continuous function need not be a bounded function. The identity function $f(x)=x$ on the real line has $|f(x)-f(y)|=|x-y|$, so $\delta =\epsilon$ always works, although its image of a function is the unbounded real line. Uniform continuity controls changes between nearby inputs, not the overall size of outputs. Likewise, an unbounded domain of a function does not by itself prevent uniform continuity.

The square root $g(x)=\sqrt{x}$ supplies a different common tolerance on the interval $[0,\infty )$. Assuming $x\ge y\ge 0$, put $u=\sqrt{x}$ and $v=\sqrt{y}$. Since $0\le u-v\le u+v$,Interchanging the inputs when necessary provesThus $\delta ={\epsilon }^2$ guarantees uniform continuity on the entire domain of a function, including zero. For $\epsilon =0.4$, any input distance below $0.16$ suffices. The pairs $(0,0.09)$ and $(4,4.09)$ both have input distance $0.09$; their output distances are respectively $0.3$ and $0.09/(\sqrt{4.09}+2)<0.0225$. The common guarantee accommodates quite different local sensitivities.

This Hölder continuity bound allows a curved tolerance rule. It is weaker than a Lipschitz bound: at zero the output change divided by the input change isNo finite Lipschitz constant works near zero, even though uniform continuity does. A Lipschitz function does have uniform continuity, choosing $\delta =\epsilon /L$ when its Lipschitz bound $L$ is positive; when $L=0$ it is a constant function.^[ref]

For general metric spaces, replace the absolute values by the respective distance functions. The Heine-Cantor theorem gives an existence guarantee: every continuous function on a compact metric space is uniformly continuous. In particular, continuity on a finite closed interval provides a common tolerance, even when no simple formula for it is known.^[ref]

For a short proof outline, fix $\epsilon >0$. Continuity gives each point $a$ an open ball of radius $r_a>0$ within which outputs differ from $f(a)$ by less than $\epsilon /2$. The open balls of half these radii still cover the domain of a function. Compactness supplies a finite subcover. Take $\delta$ to be the smallest half-radius in this finite set. Two inputs closer than $\delta$ lie together in one of the original open balls, so the triangle inequality bounds their output distance by $\epsilon$. The beautiful step is that the compact set turns local choices into finitely many choices, whose minimum stays positive.

A shared tolerance also makes uniformly continuous functions preserve Cauchy sequences: after choosing $\delta$ for $\epsilon$, all sufficiently late input pairs are closer than $\delta$, so all corresponding output pairs are closer than $\epsilon$. Unlike a convergent sequence argument, this does not require an input limit inside the domain of a function. It is why uniform continuity is useful for extending a function toward missing endpoints and for the decay argument below.

Suppose $k_1<k_2<\cdots$ are peaks of a sequence $(a_n)$ and $m_1<m_2<\cdots$ are peaks of a sequence $(-a_n)$. Then $(a_n)$ converges precisely when these two subsequences have a common finite limit of a sequence $L$.

Necessity follows because every subsequence inherits the limit of a sequence of a convergent sequence. For sufficiency, choose one selected value from each subsequence within $\epsilon$ of $L$. Every term after both selected indices lies between the two selected values:This is exactly $a_n\to L$.