What Convergence Forces
Convergence controls the terms of a sequence after some index: they must all lie close to the limit. That tail control has two important consequences. First, the entire sequence is bounded, because only finitely many terms occur before the controlled tail. Second, passing to a subsequence cannot change the limit, because the subsequence eventually uses only terms from that tail. These facts provide useful tests for proposed limits, although they do not say that every bounded sequence converges or that a convergent subsequence makes the original sequence converge.
Throughout, \((X,d)\) is a metric space, and a sequence \((x_n)\) means a term \(x_n\in X\) for every \(n\in\mathbb{N}=\{1,2,\dots\}\). We say \(x_n\to x\) if for every \(\varepsilon>0\), there is \(N\in\mathbb{N}\) such that \(d(x_n,x)<\varepsilon\) whenever \(n\ge N\). A subsequence is a sequence \((x_{n_k})\) whose indices satisfy \(n_1<n_2<\cdots\).
A Convergent Sequence Is Bounded
Proof. Suppose \(x_n\to x\) in \((X,d)\). Apply the definition of convergence with \(\varepsilon=1\). There is \(N\in\mathbb{N}\) such that \(d(x_n,x)<1\) for every \(n\ge N\). The terms before index \(N\) are finite in number. Define $$ M=\max\bigl(\{1\}\cup\{d(x_1,x),d(x_2,x),\ldots,d(x_{N-1},x)\}\bigr). $$ When \(N=1\), the second set inside the maximum is empty, so \(M=\max\{1\}=1\). In every case \(M\ge1\). For \(n<N\), the displayed definition gives \(d(x_n,x)\le M\). For \(n\ge N\), convergence gives \(d(x_n,x)<1\le M\). Thus \(d(x_n,x)\le M\) for every \(n\), so the sequence is bounded with center \(x\) and bound \(M\). \(\square\)
The proof separates the sequence into a finite initial part and a tail. The tail is bounded because it stays within distance \(1\) of the limit; the initial part is bounded because it contains only finitely many terms, each at a finite distance from the limit. This reasoning works in every metric space and does not require the space itself to be bounded.
Worked Example: Bounding a Convergent Real Sequence
In \(\mathbb{R}\) with its standard metric, consider \(x_n=2+\frac{(-1)^n}{n}\). We show directly that it converges to \(2\), find a bound, and examine two subsequences.
For every \(n\in\mathbb{N}\), the distance to \(2\) is $$ |x_n-2|=\left|\frac{(-1)^n}{n}\right|=\frac{1}{n}. $$ Given \(\varepsilon>0\), choose \(N\in\mathbb{N}\) with \(N>1/\varepsilon\). If \(n\ge N\), then \(n\ge N>1/\varepsilon\), so \(1/n<\varepsilon\). Therefore \(x_n\to2\). For a concrete check, if \(\varepsilon=1/10\), take \(N=11\): for \(n\ge11\), \(|x_n-2|=1/n\le1/11<1/10\), where \(1/11<1/10\) follows from \(10<11\).
For every \(n\ge1\), \(1/n\le1\), and hence \(|x_n-2|=1/n\le1\). Thus the sequence is bounded with center \(2\) and bound \(1\). The first terms give a check of the estimate: \(x_1=2-1=1\), \(x_2=2+1/2=5/2\), and \(x_3=2-1/3=5/3\); their distances from \(2\) are \(1\), \(1/2\), and \(1/3\), respectively.
The even-indexed subsequence has \(x_{2k}=2+\frac{1}{2k}\), so its distance from \(2\) is \(1/(2k)\), which tends to \(0\). The odd-indexed subsequence has \(x_{2k-1}=2-\frac{1}{2k-1}\), whose distance from \(2\) is \(1/(2k-1)\), also tending to \(0\). For instance, at \(k=1\) these terms are \(x_2=5/2\) and \(x_1=1\); at \(k=2\) they are \(x_4=2+1/4=9/4\) and \(x_3=2-1/3=5/3\). Both subsequences approach \(2\), in agreement with the theorem proved next.
Check: the bound uses the same center as the limit, and its inequality holds for every \(n\), including \(n=1\), where the distance is exactly \(1\). The strict inequality needed for convergence holds on the chosen tail \(n\ge11\).
Subsequences Keep the Same Limit
Proof. Suppose \(x_n\to x\), and let \((x_{n_k})\) be a subsequence, so \(n_1<n_2<\cdots\). Since the indices are strictly increasing positive integers, \(n_k\ge k\) for every \(k\): this follows by induction, because \(n_1\ge1\), and \(n_{k+1}\ge n_k+1\ge k+1\). Let \(\varepsilon>0\). Convergence of \((x_n)\) gives an \(N\) such that \(d(x_n,x)<\varepsilon\) whenever \(n\ge N\). For every \(k\ge N\), we have \(n_k\ge k\ge N\), and therefore \(d(x_{n_k},x)<\varepsilon\). This is precisely \(x_{n_k}\to x\). \(\square\)
In particular, if the original sequence converges, all of its subsequences must approach the same point. To turn this into a test, we also use uniqueness of limits in a metric space. Here is the short argument: if \(x_n\to x\) and \(x_n\to y\), then for every \(\varepsilon>0\), all sufficiently late terms satisfy both \(d(x_n,x)<\varepsilon/2\) and \(d(x_n,y)<\varepsilon/2\). The triangle inequality gives \(d(x,y)\le d(x,x_n)+d(x_n,y)<\varepsilon\). Since this holds for every \(\varepsilon>0\), \(d(x,y)=0\), and the metric axiom gives \(x=y\).
Proof. Suppose, to the contrary, that the original sequence converges to \(x\). By Theorem 2, both subsequences converge to \(x\). Each subsequence also converges to its stated limit. Uniqueness of limits then forces both stated limits to equal \(x\), contradicting that they are distinct. Therefore the original sequence does not converge. \(\square\)
Worked Example: Different Subsequences Rule Out a Limit
Let \(X=\{0,1\}\) with the discrete metric \(d(a,b)=0\) when \(a=b\) and \(d(a,b)=1\) when \(a\ne b\). Consider \(x_n=0\) for even \(n\) and \(x_n=1\) for odd \(n\).
This sequence is bounded: choosing center \(0\), its distance from \(0\) is \(0\) at every even index and \(1\) at every odd index, so \(d(x_n,0)\le1\) for every \(n\). But its even subsequence is \(x_{2k}=0\) for every \(k\), and its odd subsequence is \(x_{2k-1}=1\) for every \(k\). The constant sequence \(x_{2k}=0\) converges to \(0\), since \(d(x_{2k},0)=0<\varepsilon\) for every \(\varepsilon>0\). Likewise \(x_{2k-1}=1\) converges to \(1\), since \(d(x_{2k-1},1)=0<\varepsilon\).
These two limits are distinct because \(d(0,1)=1\). The subsequence test therefore shows that \((x_n)\) does not converge. Directly, if it converged to \(0\), the definition with \(\varepsilon=1/2\) would require all sufficiently late terms to have distance less than \(1/2\) from \(0\); every odd-indexed term instead has distance \(1\). If it converged to \(1\), every even-indexed term would have distance \(1\) from \(1\), again contradicting the required tail condition.
Check: the sequence is bounded even though it does not converge. Its terms all lie in the two-point set \(X\), and the explicit center-and-bound calculation gives \(d(x_n,0)\le1\). Boundedness is therefore a necessary consequence of convergence, not a sufficient test for it.
Using the Facts to Test Proposed Limits
The two theorems offer different kinds of evidence. If a proposed limit is \(x\), convergence requires that all sufficiently late terms be close to \(x\); in particular, every subsequence must also approach \(x\). A subsequence that stays away from \(x\), or two subsequences with different limits, rules out the proposal. Also, an unbounded sequence cannot converge, by Theorem 1. These are one-way tests: passing them does not by itself prove convergence.
Worked Example: A Proposed Limit and Its Subsequence
In \(\mathbb{R}\) with its standard metric, let \(y_n=\frac{n}{n+1}\). We test the proposed limit \(1\) and show why a subsequence approaching \(1\) does not, on its own, settle convergence to \(1\).
The distance to the proposed limit is $$ |y_n-1|=\left|\frac{n}{n+1}-1\right| =\left|\frac{n-(n+1)}{n+1}\right| =\frac{1}{n+1}. $$ Given \(\varepsilon>0\), choose \(N\in\mathbb{N}\) with \(N+1>1/\varepsilon\). For \(n\ge N\), \(n+1\ge N+1>1/\varepsilon\), so \(1/(n+1)<\varepsilon\). Thus \(y_n\to1\). The sequence is also bounded: \(0< n/(n+1)<1\) for every \(n\), since \(n>0\) and \(n<n+1\). In particular, \(|y_n-0|\le1\), so \(0\) is a center with bound \(1\).
Now take the even-indexed subsequence. Its terms are \(y_{2k}=\frac{2k}{2k+1}\), and $$ |y_{2k}-1|=\left|\frac{2k-(2k+1)}{2k+1}\right|=\frac{1}{2k+1}. $$ For \(k=1\), this gives \(y_2=2/3\) and distance \(1/3\) from \(1\); for \(k=2\), it gives \(y_4=4/5\) and distance \(1/5\). In general the distance tends to \(0\), as also follows from Theorem 2. This subsequence is consistent with the proposed limit \(1\), but that fact alone would not prove that the full sequence converges. The full-sequence calculation above supplies the needed tail estimate.
Check: substituting \(n=1\) gives \(y_1=1/2\) and \(|y_1-1|=1/2=1/(1+1)\). Substituting \(k=1\) in the subsequence formula gives \(y_2=2/3\) and \(|y_2-1|=1/3=1/(2+1)\). Both agree with the formulas used.
Common pitfall. A common incorrect argument is: “The even subsequence converges to \(0\), so the whole sequence converges to \(0\).” In the alternating sequence above, the even terms are indeed all \(0\), but every odd term is \(1\), whose distance from \(0\) is \(1\). For \(\varepsilon=1/2\), no tail can have all its terms within distance \(1/2\) of \(0\), because every tail contains odd indices. The correction is that convergence of the whole sequence forces every subsequence to share its limit; convergence of just one subsequence does not force convergence of the original sequence.
These results are useful as quick consistency checks. Before attempting a direct proof of convergence, one can ask whether the sequence is bounded and whether its natural subsequences point toward the same proposed limit. Failure of either necessary condition settles the question immediately. Passing both checks is not a proof, so a positive conclusion still requires checking the definition of convergence.
Check Your Understanding
Use the metric definition of convergence and the subsequence results to justify your answers.
- In a metric space, identify the center used in the proof that every convergent sequence is bounded. Why must the finitely many terms before the controlled tail also be included in the bound?
- Suppose \(x_n\to x\) and \(n_k\) is a strictly increasing sequence of positive integer indices. Explain why \(n_k\ge k\), and use this to prove \(x_{n_k}\to x\).
- In \(\mathbb{R}\), consider \(z_n=(-1)^n\). Find its even and odd subsequence limits and use them to decide whether \((z_n)\) converges.
- Does boundedness imply convergence in every metric space? Use the alternating sequence in the two-point discrete metric to justify your answer.
- If one subsequence of a sequence converges to a proposed point \(x\), what additional information would let you use the subsequence test to rule out convergence of the original sequence?