Limits, intermediate values and uniform extension

Learning goals and relative-domain limits

Prove intermediate values from the supremum property, use sequence criteria correctly, and extend a uniformly continuous function to a closure. If $A\subseteq\mathbb R$ and $a$ is an accumulation point of $A$, the relative-domain limit $f(x)\to L$ as $x\to a$, $x\in A$, means: every $\varepsilon>0$ has a $\delta>0$ such that $x\in A$, $0<|x-a|<\delta$ implies $|f(x)-L|<\varepsilon$. The function need not be defined at $a$.

Theorem: the sequence criterion for limits

That limit equals $L$ if and only if every sequence $x_n\in A\setminus\{a\}$ converging to $a$ has $f(x_n)\to L$.

Proof. Given the limit definition, eventually $x_n$ lies within the chosen $\delta$, so its outputs lie within $\varepsilon$. Conversely, if the limit definition fails, some $\varepsilon_0>0$ allows for each $n$ a point $x_n\in A$ with $0<|x_n-a|<1/n$ and $|f(x_n)-L|\ge\varepsilon_0$. This sequence converges to $a$ but its outputs do not converge to $L$, contradicting the stated criterion. It is important that the criterion tests every sequence, not just convenient examples.

Theorem: intermediate value property

If $f$ is continuous on $[a,b]$, $a<b$, and $f(a)<c<f(b)$, some $s\in(a,b)$ satisfies $f(s)=c$.

Proof. Let $S=\{x\in[a,b]:f(x)\le c\}$, nonempty and bounded above. Put $s=\sup S$. Continuity at $a$ and $f(a)<c$ ensure points of $S$ to the right of $a$, so $s>a$. Continuity at $b$ and $f(b)>c$ exclude a final interval near $b$ from $S$, so $s<b$. For every $n$, the supremum criterion supplies $x_n\in S$ with $s-1/n<x_n\le s$. Continuity yields $f(s)\le c$. If $f(s)<c$, continuity makes $f<c$ at points slightly to the right of $s$ still in $[a,b]$, contradicting upper boundedness by $s$. Hence $f(s)=c$. Reversing the endpoints or considering endpoint equality gives the other cases.

Uniform continuity extends to the closure

Let $f:A\to\mathbb R$ be uniformly continuous. Then there is a unique continuous extension $\overline f:\overline A\to\mathbb R$, and this extension is uniformly continuous.

Proof. For $x\in\overline A$, choose $a_n\in A$ with $a_n\to x$. This sequence is Cauchy. Uniform continuity makes $f(a_n)$ Cauchy: after choosing its one $\delta$ for a target output tolerance, sufficiently late pairs of $a_n$ are within that $\delta$. Real completeness gives a limit. If $b_n\to x$ is another choice, then $|a_n-b_n|\to0$, so uniform continuity gives $f(a_n)-f(b_n)\to0$. The limit is therefore independent of the approximating sequence; define it to be $\overline f(x)$. A constant approximating sequence at $x\in A$ proves it extends $f$.

To prove uniform continuity, fix $\varepsilon>0$ and choose $\delta>0$ for $f$ with output tolerance $\varepsilon/2$. If $x,y\in\overline A$ and $|x-y|<\delta/3$, choose sequences $a_n\to x,b_n\to y$ in $A$. For all sufficiently large $n$, $|a_n-b_n|<\delta$, so $|f(a_n)-f(b_n)|<\varepsilon/2$. Passing to limits gives $|\overline f(x)-\overline f(y)|\le\varepsilon/2<\varepsilon$, uniformly in $x,y$. Finally any continuous extension must have these same sequence limits, proving uniqueness. For $A=\varnothing$, the closure and extension are empty and the statement is vacuous.

Worked examples

A removable missing endpoint. On $(0,1)$, $f(x)=x^2$ is Lipschitz since $|x^2-y^2|\le2|x-y|$. The theorem supplies the unique continuous extension to $[0,1]$, with values zero and one at the missing endpoints.

Why ordinary continuity is insufficient. The function $1/x$ is continuous on $(0,1)$ but has no finite continuous extension at zero. Pairs $x_n=1/n$ and $y_n=1/(2n)$ for $n\ge2$ become arbitrarily close while output differences equal $n$, so it is not uniformly continuous. The theorem does not apply. Completeness of the target also matters: rational approximations can be Cauchy without converging to a rational number.

Graded practice

1. Foundation. Use the intermediate value theorem to show $x^3+x-1=0$ has a solution in $(0,1)$, then prove uniqueness.
2. Core. Show that a uniformly continuous function sends every Cauchy sequence in its domain to a Cauchy sequence.
3. Core. Give a continuous function on $(0,1)$ with no continuous extension to $[0,1]$ because of oscillation rather than unboundedness.
4. Proof. If two continuous functions on $\mathbb R$ agree on $\mathbb Q$, prove they agree everywhere. Does agreement on the integers suffice?

Full solutions

1. The polynomial takes values $-1$ and $1$ at the endpoints and is continuous, so an intermediate zero exists. If $y>x$, then $(y^3+y)-(x^3+x)=(y-x)(y^2+xy+x^2+1)>0$, so the function is strictly increasing and has at most one zero.
2. Given $\varepsilon>0$, choose a uniform $\delta$ for the function. The Cauchy input sequence has every pair of sufficiently late terms within $\delta$, hence their outputs differ by less than $\varepsilon$. This is exactly the output Cauchy condition.
3. Take $f(x)=\sin(1/x)$. Sequences $x_n=1/(\pi/2+2n\pi)$ and $y_n=1/(3\pi/2+2n\pi)$ tend to zero but have outputs $1$ and $-1$. Thus no endpoint limit exists despite boundedness.
4. For any real $x$, rational density gives a sequence $q_n\to x$. Continuity gives $f(x)=\lim f(q_n)=\lim g(q_n)=g(x)$. The integers are not dense: $f(x)=0$ and $g(x)=\sin(\pi x)$ agree at every integer but not at $1/2$.

Summary and term definitions

An accumulation point is approached by other domain points. A closure adds all limit points. A continuous extension fills missing values without breaking continuity. Uniform continuity controls pairs across the whole domain and thereby preserves Cauchy behavior; target completeness supplies the limits needed to fill the closure. Intermediate values follow from completeness plus continuity, not from a graph's appearance.