This is an interesting observation made by one of my friends. After thinking for a bit, I think I was able to find the solution which I will write as an answer. I feel this is worth posting, both because it's interesting and perhaps pedagogically useful. For simplicitly, we will work with $\limsup$ but $\liminf$ is basically the same thing.

For any non-empty, bounded $A,B \subset \mathbb{R}$, we define $P = \{x + y : x \in A, y \in B\}$. It's easy to see that $\sup P = \sup A + \sup B$. Therefore, \begin{align}\lim \sup a_n + \lim \sup b_n &= \lim_{m \to \infty} \sup_{n \geq m} a_n + \lim_{m \to \infty} \sup_{n \geq m}b_n = \lim_{m \to \infty} \left(\sup_{n \geq m} a_n + \sup_{n \geq m}b_n\right) \\[0.3cm]&= \lim_{m \to \infty} \left(\sup_{n \geq m} \left(a_n + b_n\right)\right) = \lim \sup \ (a_n + b_n)\end{align}

This is wrong, however. The relationship between the left-most expression and the right-most is $\geq$. Where is the mistake?


1 Answer 1


For sequences, it does not hold that $$\sup_{n \geq m} a_n + \sup_{n \geq m}b_n = \sup_{n \geq m} \left(a_n + b_n\right) \ \ \ (1)$$

For example, fix $m$ and consider the sequence $a_n = \sin(\frac{\pi n}{2})$ and $b_n = -a_n$. Considering the above equation, the right hand side is $0$ and the left hand side is $2$.

Well, why does it hold for sets and not sequences? Well, for sets, there is no order.

In the question, the set $P$ is defined as $\{x + y : x \in A, y \in B\}$. This set encompasses all of the possible ways one can add elements of $A$ to elements of $B$. However, for sequences there is the "restriction" that you can only add $a_i$ to $b_i$. Hence the "large" values in one sequence may not be paired up with the "large" values of the other sequence.

Let's go back to the above example $a_n = \sin(\frac{\pi n}{2})$, $b_n = -a_n$. With a set, we would consider all $6$ ways we could add two elements from the set $\{0, 1, -1 \}$ and take the $\sup$ of that. On the other hand, with sequences, we may not be able to consider all $6$ ways we could add two elements from the above set because they don't all occur in the sequence $a_n + b_n$; in fact, only one does, namely $0 + 0$.

Thus, what we can say is that $$\sup_{n \geq m} a_n + \sup_{n \geq m}b_n = \sup \{a_k + b_j : j,k \in \mathbb{N} \ \text{and} \ k,j \geq m \} \geq \sup_{n \geq m} \left(a_n + b_n\right)$$

since $$\{a_n + b_n : n \geq m\} \subset \{a_k + b_j : j,k \in \mathbb{N} \ \text{and} \ k,j \geq m \}$$

It's not hard to show that $(1)$ holds if $a_n$ and $b_n$ are monotone increasing. This is a sufficient condition. I'm not sure of a necessary and sufficient condition.

  • $\begingroup$ There's no such thing as the supremum of a sequence, it's always about sets one way or another. $\endgroup$
    – Git Gud
    Oct 14, 2016 at 8:12
  • $\begingroup$ @GitGud A sequence is a function $f:\mathbb{N} \to \mathbb{R}$. Its supremum can be defined as the supremum of its image. I couldn't think of anything better to call it in this context. $\endgroup$ Oct 14, 2016 at 8:16
  • $\begingroup$ On that we agree. But I disagree with the use of terminology in "Well, why does it hold for sets and not sequences?". $\endgroup$
    – Git Gud
    Oct 14, 2016 at 8:18

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