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Assuming that $a_{n}$ and $b_{n}$ are two bounded sequences of real numbers, I need to prove that $\inf(a_{n} + b_{n}) \leq \inf(a_{n}) + \sup(b_{n})$.

I have seen proofs that that $\inf(a_{n}+b_{n}) \leq \inf(a_{n}) + \inf(b_{n})$ through this post. Am I missing something obvious here? I'm having a hard time finding an example for the inequality to hold true.

Giorgos Giapitzakis
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gus f
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2 Answers2

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As pointed out by GEdgar, the inequality $\inf(a_{n}+b_{n}) \leq \inf(a_{n}) + \inf(b_{n})$ may fail, for example if $a_n=(-1)^n$ and $b_n=-a_n$.

In order to get the inequality you want, let $B=\sup(b_k)$. Then for each $n$, $a_n+b_n\leqslant a_n+B$. Take the infimum and find a link between $\inf(a_n+B)$, $B$ and $\inf(a_n)$.

Davide Giraudo
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Assume on the contrary that $\inf(a_{n}+b_{n})>c>\inf(a_{n})+\sup(b_{n})$. $(1)$

Now take a subsequence of $a_{n}$, say $a_{n_{k}}\to\,\inf(a_{n})$.

By $(1)\,\,\, a_{n_{k}}+b_{n_{k}}>c>\inf(a_{n})+\sup(b_{n})$. Taking $\liminf$

we get $\inf(a_{n})+\liminf\,b_{n_{k}}\,\geq\,c\,>\inf(a_{n})+\sup(b_{n})$

which implies $\liminf\,b_{n_{k}}\,>\,\sup(b_{n})$ which implies that there is a further subsequence $b_{n_{k_{m}}}$

converging to a number greater than $\sup(b_{n})$ , contradiction.

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    Are you aware that (almost) all mathematical functions have corresponding commands in MathJax/LaTeX? Try \inf, \sup, \liminf etc, that looks much better? – Details at https://math.meta.stackexchange.com/q/5020/42969. – Martin R Sep 15 '22 at 15:02
  • Thanks I will use them ! –  Sep 15 '22 at 16:05
  • Note that there need not be a subsequence of $a_n$ converging to $\inf a_n$. This holds iff $\inf a_n = \liminf a_n$ but in general we only know $\inf a_n \leq \liminf a_n$. Arguably the remaining case is much simpler (nearly trivial if you choose good cases), but at least splitting into cases would make choosing that subsequence valid. – Brian Moehring Sep 17 '22 at 17:44
  • Brian, this is true for all sets not just for sequences! If $infA=c$ then there is a sequence in A converging to c. Because taking $(c,c+\dfrac{1}{n})$ there must be an element of A in $(c,c+\dfrac{1}{n})$. Otherwise $c+\dfrac{1}{n}$ would be a lower bound greater than c, contradiction! It has nothing to do with liminf. –  Sep 17 '22 at 17:53
  • @GeorgeTsoutsinos Take, say, $a_1 = -1$, $a_n = 0$ for all $n \geq 2$. What subsequence of ${ a_n }$ do you propose that converges to $\inf a_n = -1$? (This of course being a simple example in the spirit of Brian's comment.) – Jakob Streipel Sep 17 '22 at 20:46
  • Yes ! I agree ! Your argument is correct. But my statement is true when inf is an inf and not a min. If inf$(a_{n})=a_{M}$ then my proof gets even more simpler, because we have inf$(a_{n}+b_{n})$>$c$>$,a_{M}$+sup$(b_{n})$ where $a_{M}$ is your min and we get $a_{M}$+$b_{M}$>$a_{M}$+sup$(b_{n})$ giving $b_{M}$>sup$(b_{n})$ contradiction. I just chose the difficult case and not the simple case. But your point is correct! In general (or usually) we assume that there are points of the set arbitrarily close to the inf . –  Sep 17 '22 at 21:09