When Can Terms Be Chosen to Have One Fixed Value?
A subsequence is formed by selecting terms at strictly increasing indices. The selection can sometimes produce a particularly simple sequence: every selected term may have the same value. For example, an alternating sequence can have a constant subsequence even though the original sequence itself is not constant.
The key question is whether some value appears at infinitely many indices. If so, those indices can be selected in increasing order, giving a constant subsequence. If every value occurs only finitely many times, no constant subsequence is possible. This gives an exact criterion, not merely a useful sufficient condition.
The constant value need not be \(a_0\), and the original sequence need not be constant. The defining requirements are that the value stay fixed along the selected terms and that the selected indices increase strictly.
The Exact Criterion: A Value Occurs Infinitely Often
For a fixed real number \(c\), the indices at which a sequence takes the value \(c\) form the set
Saying that \(c\) occurs infinitely often means exactly that \(E_c\) is infinite. The construction method from “Constructing Subsequences” can then select an increasing sequence of indices from \(E_c\). Conversely, a constant subsequence supplies infinitely many distinct indices at which its constant value occurs.
Proof. First suppose that \((a_n)\) has a constant subsequence \((a_{n_k})\) with constant value \(c\). Then \(a_{n_k}=c\) for every \(k\). The indices \(n_0,n_1,\ldots\) are strictly increasing, so they are all distinct. Therefore \(c\) occurs at infinitely many indices in the original sequence.
Now suppose there is a real number \(c\) such that \(E_c=\{n\in\mathbb{N}_0:a_n=c\}\) is infinite. By the theorem “Constructing a Subsequence from Infinitely Many Eligible Indices,” there is a subsequence \((a_{n_k})\) whose indices all belong to \(E_c\). Thus \(a_{n_k}=c\) for every \(k\), so this subsequence is constant. Both implications hold, proving the criterion. \(\square\)
The theorem identifies exactly what must be checked: not whether terms repeat at least once, or even a large but finite number of times, but whether one fixed value occurs infinitely often. The infinite set of matching indices is what makes an infinite subsequence possible.
Worked Example: A Constant Subsequence in an Alternating Sequence
Define \(a_n=3\) when \(n\) is even and \(a_n=-4\) when \(n\) is odd. For each \(k\in\mathbb{N}_0\), choose \(n_k=2k\). These indices are strictly increasing because
Every selected index is even, so \(a_{n_k}=3\) for every \(k\). In particular, \(a_0=3\), \(a_2=3\), and \(a_4=3\), and the same equality holds for all later selected terms. Thus \((a_{2k})_{k=0}^{\infty}\) is a constant subsequence with value \(3\). The original sequence is not constant, since \(a_0=3\) and \(a_1=-4\).
Finite Range Guarantees a Constant Subsequence
A sequence has finite range if all its terms belong to some finite set of real numbers. For example, a sequence whose terms are always among four prescribed values has a finite range. The terms may switch among those values in an irregular way, but there are only finitely many possible values available to occur infinitely often.
Proof. Let \(F\) be the finite, nonempty set of values taken by the sequence. For each \(c\in F\), let \(E_c=\{n\in\mathbb{N}_0:a_n=c\}\). Every index belongs to one of these sets, so
The set \(\mathbb{N}_0\) is infinite, and this is a cover by finitely many sets. By the “Finite Partition Principle for Infinite Sets” from “Constructing Subsequences,” at least one of the sets \(E_c\) is infinite. For that value \(c\), the criterion for a constant subsequence applies: there is a constant subsequence with value \(c\). \(\square\)
This argument does not require the sequence to cycle through its values in a regular pattern. It uses only two facts: the indices are infinite, and finitely many value-classes cover all of them. One class must contain infinitely many indices.
Worked Example: A Sequence Taking Four Values
For each \(n\in\mathbb{N}_0\), write \(n=4q+r\), where \(q\in\mathbb{N}_0\) and \(r\in\{0,1,2,3\}\), and define \(a_n=r\). Thus the values repeat in the pattern \(0,1,2,3\). The indices \(n_k=4k\) are strictly increasing, since
At each such index, \(4k=4k+0\), so the remainder is \(0\) and \(a_{n_k}=0\). Hence \((a_{4k})_{k=0}^{\infty}\) is a constant subsequence. This explicit choice exhibits the repeated value directly; the finite-range theorem guarantees that some constant subsequence exists even when the pattern is not so easy to see.
Repeated Values Need Not Follow a Simple Pattern
The infinitely repeated value may occur at indices that are not consecutive and do not have a fixed difference. The criterion still applies: once the matching index set is known to be infinite, the construction theorem selects its elements in increasing order.
Worked Example: A Constant Value at Square Indices
Define a sequence by setting \(a_n=7\) if \(n=k^2\) for some \(k\in\mathbb{N}_0\), and \(a_n=-n\) otherwise. The square indices can be listed as \(n_k=k^2\). They increase strictly because
for every \(k\in\mathbb{N}_0\). By the definition of the sequence, \(a_{n_k}=a_{k^2}=7\) for every \(k\). For the first few indices, \(n_0=0\), \(n_1=1\), and \(n_2=4\), giving \(a_0=a_1=a_4=7\). Therefore the selected terms form a constant subsequence, even though the indices become farther apart.
This example also illustrates why a constant subsequence is a selection of terms, not a claim that every term between selected indices has the same value. At indices that are not squares, the sequence instead takes the value \(-n\). Those intervening terms do not affect the selected subsequence.
Boundedness Alone Does Not Guarantee a Constant Subsequence
A common overreach is to assume that a sequence has a constant subsequence merely because its terms stay within fixed bounds. The criterion shows why that is not enough: boundedness does not force any one exact value to occur infinitely often. A sequence can have infinitely many distinct values while remaining bounded.
Worked Example: A Bounded Sequence with No Constant Subsequence
Consider \(a_n=\dfrac{1}{n+2}\) for \(n\in\mathbb{N}_0\). Every term satisfies \(0<a_n\leq\dfrac12\), so the sequence is bounded. If \(m<n\), then
because \(n-m>0\) and both factors in the denominator are positive. Thus \(a_m\neq a_n\) whenever \(m\neq n\): all terms are distinct. No value can occur at infinitely many indices, so the criterion proves that the sequence has no constant subsequence. This does not rule out other kinds of subsequences; it establishes specifically that no subsequence can have one fixed value throughout.
There is also an important distinction between a constant subsequence and a sequence that is eventually constant. A constant subsequence only uses selected indices, which can leave out many other terms. For instance, the alternating sequence above has a constant subsequence, although its terms continue to alternate at every step and it is not eventually constant.
When looking for a constant subsequence, it is useful to separate the task into two questions. First, can one show that a particular value occurs infinitely often? If so, choose its indices in increasing order. If the sequence takes only finitely many possible values, the finite-range principle supplies such a value without requiring it to be identified in advance. If the range is not finite, the criterion remains valid, but a separate argument is needed to establish an infinitely repeated value.
Choose \(c\), or list the finitely many possible values if the sequence has finite range.
Write \(E_c=\{n\in\mathbb{N}_0:a_n=c\}\) and establish that this set is infinite.
Use the construction theorem to choose an increasing sequence of indices from \(E_c\).
Verify that each selected index belongs to \(E_c\), so every selected term equals \(c\).
The precise criterion prevents two opposite errors. A few repeated terms do not establish a constant subsequence, because the value must occur infinitely often. On the other hand, the indices need not be consecutive or regularly spaced: infinitude of the matching indices is enough to select a subsequence.
Check Your Understanding
Use the criterion and the finite-range principle to answer the following questions.
- For a fixed real number \(c\), what does it mean to say that \(c\) occurs infinitely often in \((a_n)\)?
- Why does a constant subsequence imply that its constant value occurs at infinitely many indices of the original sequence?
- Which earlier result guarantees a constant subsequence when a sequence takes only finitely many values?
- Can a bounded sequence fail to have a constant subsequence? Give a reason based on the criterion.
- Does having a constant subsequence imply that the original sequence is eventually constant? Explain.