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For a short proof outline, endpoint values give the non-strict cases immediately. If $y$ is strictly between $f(a)$ and $f(b)$, apply the bisection method to the continuous function $g(x)=f(x)-y$. A midpoint with $g=0$ finishes the proof. Otherwise retain nested intervals $[a_k,b_k]$ with strictly opposite signs and widths $(b-a)2^{-k}$. The nested interval theorem, using completeness of the real numbers, gives a common point $c$ to which both endpoint sequences converge. After reversing the sign of $g$ if necessary, $g(a_k)<0<g(b_k)$. Continuity makes both endpoint value sequences converge to $g(c)$, forcing $g(c)\le 0$ and $g(c)\ge 0$, hence $f(c)=y$. The original endpoint values differ from $y$, so $c\in (a,b)$. This proof explains why a jump can defeat the construction: shrinking a candidate closed interval cannot supply the missing continuity.

A sequence of nested intervals keeps every new enclosure inside the previous one:Each step adds a restriction without discarding an earlier restriction. This is useful when an exact real number is described by increasingly precise bounds rather than by a finite decimal expansion. The bisection method, for instance, produces nested intervals that keep enclosing a zero of a function.

For a simple example, the closed intervals $I_n=[1-1/n,1+1/n]$ are nested and all contain $1$. Their lengths $2/n$ tend to zero, so the nested interval theorem leaves exactly that point. The family $J_n=[1-1/n,2+1/n]$ is also nested, but its intersection is the whole closed interval $[1,2]$. Nesting alone does not force uniqueness.

The open intervals $K_n=(0,1/n)$ illustrate a different possibility. Every finite collection has a nonempty intersection, but the intersection of all of them is empty: any positive candidate eventually exceeds $1/n$, and zero is excluded from every member. Each finite row in the diagram contains points; the full infinite family need not.

Begin with a closed interval $I_0=[a,b]$ containing the entire bounded sequence. Split the interval at its midpoint. At least one of the two closed halves contains infinitely many indexed terms; choose such a half and repeat. This gives nested intervals $I_j$ with lengths $(b-a)/2^j$. The nested interval theorem, using completeness of the real numbers, supplies their unique common point $\ell$.

Choose $n_j>n_{j-1}$ with $x_{n_j}\in I_j$. Infinitely many indexed terms in $I_j$ guarantee that the next index can be chosen. Both $x_{n_j}$ and $\ell$ belong to $I_j$, soThis is a short existence proof, not necessarily a finite algorithm for an arbitrary input sequence: deciding which half contains infinitely many future terms may require mathematical knowledge about that sequence.

The set $I^n$ consists precisely of points whose first $n$ positions $x,f(x),\dots ,f^{n-1}(x)$ all lie in $L\cup R$. HenceDeleting the first position preserves this condition, proving $f(\Lambda )\subseteq \Lambda$. For the reverse inclusion, if $y\in \Lambda$, take $x=g_L(y)$. Then $x\in L$ and $f^k(x)=f^{k-1}(y)\in L\cup R$ for every $k\ge 1$. Thus $x\in \Lambda$ and $f(x)=y$. In fact both inverse branches give preimages in $\Lambda$. Thereforewhich proves full invariance, not merely forward invariance.

For any infinite symbolic sequence $s$, the itinerary cylinders $I_{s_1\cdots s_n}$ are nonempty compact nested intervals. The nested interval theorem gives at least one point in their intersection. Every $x\in \Lambda$ has a unique itinerary because $L\cap R=\varnothing$. To justify that the itinerary also determines a unique point, it suffices to show the interval lengths shrink. The fact that logistic inverse branches contract in an angular coordinate works for every $\mu >4$; a direct claim that $|g_L^{\prime }|,|g_R^{\prime }|<1$ in the original coordinate would fail for some $\mu$ just above four.

Use the homeomorphism $h(u)={\sin }^2(\pi u/2)$ of $[0,1]$. The transformed inverse branches areTheir derivatives satisfyThe inequality follows by squaring and using $4/\mu <1$. Thus every length-$n$ transformed itinerary cylinder has length at most ${\mu }^{-n/2}$. Since $|h^{\prime }|\le \pi /2$, its original interval has length at most $(\pi /2){\mu }^{-n/2}$ and shrinks to a point. This gives the claimed one-to-one identification with the one-sided full shift.

Finally part (ii) givesThus $f$ acts by the left shift. The identification is a topological conjugacy: matching long prefixes forces closeness by the shrinking estimate, while distinct finite itinerary cylinders are disjoint compact sets and so their inverse coding is continuous. In particular $f$ need not be one-to-one on $\Lambda$, just as a one-sided left shift has two possible first symbols before any given tail.