Let \(f:[0,\infty)\to\mathbb R\) be continuous. Let \(A\) be the set of all real numbers \(a\) for which there exists a sequence \(x_n\to\infty\) such that \[ f(x_n)\to a. \] Prove that if \(a,b\in A\), then every point of the interval with endpoints \(a\) and \(b\) also belongs to \(A\).
Solution
1. It is enough to consider the case\[ a<b \]and a point\[ c\in(a,b). \]
If \(a=b\), the interval with endpoints \(a\) and \(b\) is the singleton \(\{a\}\subseteq A\). If \(a\neq b\), interchange the names of \(a\) and \(b\) if necessary so that \(a<b\). The endpoints already lie in \(A\) by hypothesis, so only the points strictly between them require proof.
2. There are sequences \((u_n)\) and \((v_n)\) in \([0,\infty)\) such that\[ u_n\geq n, \qquad v_n\geq n, \]and\[ f(u_n)<c<f(v_n) \]for every \(n\).
Because \(a\in A\), choose a sequence \(x_j\to\infty\) with \(f(x_j)\to a\). Since \(a<c\), for every sufficiently large \(j\) one has \(f(x_j)<c\). Since also \(x_j\to\infty\), for each \(n\) one may choose an index \(j_n\) large enough that \[ x_{j_n}\geq n \qquad\text{and}\qquad f(x_{j_n})<c. \] Set \(u_n\coloneqq x_{j_n}\).
Similarly, since \(b\in A\) and \(c<b\), choose \(y_j\to\infty\) with \(f(y_j)\to b\) and, for each \(n\), an index \(k_n\) such that \[ y_{k_n}\geq n \qquad\text{and}\qquad f(y_{k_n})>c. \] Set \(v_n\coloneqq y_{k_n}\).
3. For every \(n\), there is \(z_n\geq n\) such that\[ f(z_n)=c. \]
The function \(f\) is continuous on the closed interval whose endpoints are \(u_n\) and \(v_n\). By step 2, its values at those endpoints lie on opposite sides of \(c\). The intermediate value theorem therefore gives a point \(z_n\) between \(u_n\) and \(v_n\) with \(f(z_n)=c\). Since both endpoints are at least \(n\), one has \(z_n\geq n\).
4. The point \(c\) belongs to \(A\).
Step 3 gives \(z_n\geq n\), hence \(z_n\to\infty\), while \[ f(z_n)=c \] for every \(n\). Thus \(f(z_n)\to c\), which is exactly the condition \(c\in A\).
5. Q.E.D.
By step 1, the endpoints need no proof, and step 4 gives every strict intermediate point. Hence the entire interval with endpoints \(a\) and \(b\) lies in \(A\).