Course Material 2.8 Subsequences and the Bolzano- Weierstrass Theorem
By a subsequence of a given sequence {𝑥𝑛 }, we mean a sequence formed by
taking terms from {𝑥𝑛 } and arranging them in the same order as in {𝑥𝑛 }. For example, if
we take only the even terms of {𝑥𝑛 } we get a sub sequence 𝑥2 , 𝑥4 , 𝑥6 , … 𝑥2𝑛 , … …
Taking the odd terms, we get the sub sequence 𝑥1 , 𝑥3 , 𝑥5 , … 𝑥2𝑛−1 , … …. These two
subsequences may be taken as {𝑥2𝑘 } and {𝑥2𝑘+1 }. They are called the even sub sequence
and the odd subsequence of {𝑥𝑛 }.
Hence going for a definition of a subsequence we observe that a sub sequence of
a given sequence {𝑥𝑛 }, is a sequence of the form {𝑥𝑛𝑘 } for 𝑘 = 1,2,3, … ..where 𝑛1 < 𝑛2 <
𝑛3 < ⋯ < 𝑛𝑘 < 𝑛𝑘+1 , … … or 𝑛𝑘 is a strictly increasing sequence of positive integers.
Clearly, for the even subsequence, 𝑛𝑘 = 2𝑘
and for the odd sub sequence, 𝑛𝑘 = 2𝑘 − 1
Both are strictly increasing functions of k
Definition.
Given a sequence {𝑥𝑛 } and if {𝑛𝑘 } is a strictly increasing sequence of positive integers
then then the sequence {𝑥𝑛𝑘 } is said to be a subsequence of {𝑥𝑛 }.
Note:
1. For all 𝑘 we have 𝑘 ≤ 𝑛𝑘
2. If 𝑘 > 𝑖 we have 𝑛𝑘 > 𝑛𝑖
3. If 𝑛𝑘0 > 𝑁 then 𝑛𝑘 > 𝑁 for all 𝑘 ≥ 𝑘0
4. A subsequence {𝑥𝑛𝑘 } of {𝑥𝑛 } is said to converge to a real number 𝑥 if given 𝜖 > 0
there exists a positive integer 𝑘0 such that |𝑥𝑛𝑘 − 𝑥| < 𝜖 for all 𝑘 ≥ 𝑘0 .
5. Trivially, the sequence {𝑥𝑛 } itself is a subsequence by taking 𝑛𝑘 = 𝑘 for all 𝑘 ∈ ℕ
, Theorem
If a sequence {𝑥𝑛 } converge to a real number 𝑥, then all its subsequences will
converge to 𝑥
Proof
Suppose {𝑥𝑛 } converge to 𝑥. Then given 𝜖 > 0 there exists a positive integer 𝑁 such
that |𝑥𝑛 − 𝑥 | < 𝜖 for all 𝑛 ≥ 𝑁 .
Let {𝑥𝑛𝑘 } be a subsequence of {𝑥𝑛 }. Since 𝑛𝑘 is a strictly increasing sequence of
positive integers, we can find a positive integer 𝑘0 such that 𝑛𝑘0 > 𝑁. Then for all
𝑘 ≥ 𝑘0 , we have 𝑛𝑘 > 𝑁 and hence |𝑥𝑛𝑘 − 𝑥| < 𝜖, thus proving that the sub
sequence {𝑥𝑛𝑘 } converges to 𝑥.
Remark.
The converse of the above result is not true. For example, for the sequence {(−1)𝑛+1 },
the odd subsequence is the constant sequence 1,1,1,…….. which converges to 1. But the
sequence is not convergent. The even subsequence is also convergent but it converges
to a different limit. But the converse will be true if {𝑥𝑛 } is a Cauchy sequence.
Theorem
If {𝑥𝑛 } is a Cauchy sequence and if a subsequence {𝑥𝑛𝑘 } converges to 𝑥 then {𝑥𝑛 } will
also converge to 𝑥
Proof :
Suppose that {𝑥𝑛 } is a Cauchy sequence. Then given 𝜖 > 0, we can find a positive integer
𝜖
𝑁1 , such that for all 𝑛 ≥ 𝑁1, we have |𝑥𝑛 − 𝑥𝑚 | < 2 ---(1)
Assume also that there is a sub sequence {𝑥𝑛𝑘 } that converges to the real number 𝑥.
𝜖
Then there exists a positive integer 𝑘0 such that |𝑥𝑛𝑘 − 𝑥| < 2 for all 𝑘 ≥ 𝑘0 .---(2)
𝜖
As 𝑛𝑘 is strictly increasing, we can choose 𝑘0 such that 𝑛𝑘0 ≥ 𝑁 so that |𝑥𝑛𝑘0 − 𝑥|<2
𝜖 𝜖
Then for all 𝑛 ≥ 𝑁 |𝑥𝑛 − 𝑥 | ≤ |𝑥𝑛 − 𝑥𝑛𝑘0 | + |𝑥𝑛𝑘0 − 𝑥| < + =𝜖
2 2
This shows that {𝑥𝑛 } converges to 𝑥
By a subsequence of a given sequence {𝑥𝑛 }, we mean a sequence formed by
taking terms from {𝑥𝑛 } and arranging them in the same order as in {𝑥𝑛 }. For example, if
we take only the even terms of {𝑥𝑛 } we get a sub sequence 𝑥2 , 𝑥4 , 𝑥6 , … 𝑥2𝑛 , … …
Taking the odd terms, we get the sub sequence 𝑥1 , 𝑥3 , 𝑥5 , … 𝑥2𝑛−1 , … …. These two
subsequences may be taken as {𝑥2𝑘 } and {𝑥2𝑘+1 }. They are called the even sub sequence
and the odd subsequence of {𝑥𝑛 }.
Hence going for a definition of a subsequence we observe that a sub sequence of
a given sequence {𝑥𝑛 }, is a sequence of the form {𝑥𝑛𝑘 } for 𝑘 = 1,2,3, … ..where 𝑛1 < 𝑛2 <
𝑛3 < ⋯ < 𝑛𝑘 < 𝑛𝑘+1 , … … or 𝑛𝑘 is a strictly increasing sequence of positive integers.
Clearly, for the even subsequence, 𝑛𝑘 = 2𝑘
and for the odd sub sequence, 𝑛𝑘 = 2𝑘 − 1
Both are strictly increasing functions of k
Definition.
Given a sequence {𝑥𝑛 } and if {𝑛𝑘 } is a strictly increasing sequence of positive integers
then then the sequence {𝑥𝑛𝑘 } is said to be a subsequence of {𝑥𝑛 }.
Note:
1. For all 𝑘 we have 𝑘 ≤ 𝑛𝑘
2. If 𝑘 > 𝑖 we have 𝑛𝑘 > 𝑛𝑖
3. If 𝑛𝑘0 > 𝑁 then 𝑛𝑘 > 𝑁 for all 𝑘 ≥ 𝑘0
4. A subsequence {𝑥𝑛𝑘 } of {𝑥𝑛 } is said to converge to a real number 𝑥 if given 𝜖 > 0
there exists a positive integer 𝑘0 such that |𝑥𝑛𝑘 − 𝑥| < 𝜖 for all 𝑘 ≥ 𝑘0 .
5. Trivially, the sequence {𝑥𝑛 } itself is a subsequence by taking 𝑛𝑘 = 𝑘 for all 𝑘 ∈ ℕ
, Theorem
If a sequence {𝑥𝑛 } converge to a real number 𝑥, then all its subsequences will
converge to 𝑥
Proof
Suppose {𝑥𝑛 } converge to 𝑥. Then given 𝜖 > 0 there exists a positive integer 𝑁 such
that |𝑥𝑛 − 𝑥 | < 𝜖 for all 𝑛 ≥ 𝑁 .
Let {𝑥𝑛𝑘 } be a subsequence of {𝑥𝑛 }. Since 𝑛𝑘 is a strictly increasing sequence of
positive integers, we can find a positive integer 𝑘0 such that 𝑛𝑘0 > 𝑁. Then for all
𝑘 ≥ 𝑘0 , we have 𝑛𝑘 > 𝑁 and hence |𝑥𝑛𝑘 − 𝑥| < 𝜖, thus proving that the sub
sequence {𝑥𝑛𝑘 } converges to 𝑥.
Remark.
The converse of the above result is not true. For example, for the sequence {(−1)𝑛+1 },
the odd subsequence is the constant sequence 1,1,1,…….. which converges to 1. But the
sequence is not convergent. The even subsequence is also convergent but it converges
to a different limit. But the converse will be true if {𝑥𝑛 } is a Cauchy sequence.
Theorem
If {𝑥𝑛 } is a Cauchy sequence and if a subsequence {𝑥𝑛𝑘 } converges to 𝑥 then {𝑥𝑛 } will
also converge to 𝑥
Proof :
Suppose that {𝑥𝑛 } is a Cauchy sequence. Then given 𝜖 > 0, we can find a positive integer
𝜖
𝑁1 , such that for all 𝑛 ≥ 𝑁1, we have |𝑥𝑛 − 𝑥𝑚 | < 2 ---(1)
Assume also that there is a sub sequence {𝑥𝑛𝑘 } that converges to the real number 𝑥.
𝜖
Then there exists a positive integer 𝑘0 such that |𝑥𝑛𝑘 − 𝑥| < 2 for all 𝑘 ≥ 𝑘0 .---(2)
𝜖
As 𝑛𝑘 is strictly increasing, we can choose 𝑘0 such that 𝑛𝑘0 ≥ 𝑁 so that |𝑥𝑛𝑘0 − 𝑥|<2
𝜖 𝜖
Then for all 𝑛 ≥ 𝑁 |𝑥𝑛 − 𝑥 | ≤ |𝑥𝑛 − 𝑥𝑛𝑘0 | + |𝑥𝑛𝑘0 − 𝑥| < + =𝜖
2 2
This shows that {𝑥𝑛 } converges to 𝑥
