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I realize there are already other duplicates, but none of them are really my question.

In the proof for $\sup (A+B) = \sup A + \sup B$; why is it sufficient to show that $\sup (A +B) + \epsilon \geq \sup A + \sup B$ for all $\epsilon >0$. I already understand the other direction with $\sup (A+B) \leq \sup A + \sup B$

Lemon
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6 Answers6

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If $$\sup(A+B)<\sup A+\sup B,$$ then there is some $\epsilon>0$ such that $$\sup(A+B)+\epsilon<\sup A+\sup B.$$ Thus, in showing that $$\sup(A+B)+\epsilon\ge\sup A+\sup B$$ for all $\epsilon>0,$ we show that $$\sup(A+B)\ge\sup A+\sup B$$ by contrapositive.

Edit: For a moment, let's forget about the $\epsilon$, and deal with a statement that should be less obscure:

Given real numbers $x,y$. If every real number bigger than $x$ is at least as big as $y$, then $x$ is at least as big as $y$. Symbolically, $$\bigl[\forall z>x\:(z\ge y)\bigr]\Longrightarrow(x\ge y).\tag{$\heartsuit$}$$

Again, we prove this by contrapositive. Suppose $x<y$. Then there is some $x<z<y$. (In particular, letting $z=\frac{x+y}2$ be the midpoint of $x,y$ does the trick.) $\Box$

(The converse of the above holds readily by transitivity of the real order relation.)

Now, $$z>x\quad\Longleftrightarrow\quad z-x>0$$ and $$z\ge y\quad\Longleftrightarrow\quad x+(z-x)\ge y.$$ Letting $\epsilon:=z-x,$ we then have $z>x$ if and only if $\epsilon>0$, and $z\ge y$ if and only if $x+\epsilon\ge y.$ Thus, $(\heartsuit)$ can be rewritten as $$\bigl[\forall\epsilon>0\:(x+\epsilon\ge y)\bigr]\Longrightarrow(x\ge y),\tag{$\star$}$$ or, in words, if $x+\epsilon$ is at least as big as $y$ for all positive real numbers $\epsilon$, then $x$ is at least as big as $y$.

As a further alternative, put into the language of sets and subsets, $(\star)$ can be phrased: "If $(x,\infty)\subseteq[y,\infty),$ then $[x,\infty)\subseteq[y,\infty).$"

Community
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Cameron Buie
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  • may I prove $\star$ as follows? $y \leq x \implies y + \epsilon \leq x + \epsilon; y \leq y + \epsilon \leq x + \epsilon$. On the other hand, we have $\lim_{\epsilon \to 0} y \leq \lim_{\epsilon \to 0} (x + \epsilon) = x$ – Lemon Jun 23 '13 at 22:11
  • @sidht: Your first line looks like you're assuming what you want to prove. Can you clarify what you're assuming? – Cameron Buie Jun 23 '13 at 22:36
  • I want to prove $y \leq x \iff y \leq x + \epsilon$. Or I guess we could turn the inequality on the RHS to a strict inequality. – Lemon Jun 23 '13 at 22:42
  • Ah! I see. Indeed, then $y\le x\implies y\le x+\epsilon$ for all $\epsilon>0$, exactly as you described. You'll need to be a bit cautious about using limits to go the other way, since our definition of limits already uses an $\epsilon$, and uses $x$ as a variable! It might be better to proceed as follows: (1) Note that $0$ is a lower bound of the set of positive reals (by definition of positive). (2) Note/prove that half of a positive number is a smaller positive number, thus proving that $0$ is the infimum of the set of positive reals. (3) Suppose $y\le x+\epsilon$ for all $\epsilon>0.$ ... – Cameron Buie Jun 23 '13 at 22:54
  • ... But that means $y-x\le\epsilon$ for all $\epsilon>0,$ meaning $y-x$ is a lower bound of the positive numbers, so since $0$ is the greatest lower bound of the positive numbers, we have $y-x\le 0,$ and so $y\le x$. – Cameron Buie Jun 23 '13 at 22:55
  • Sorry I am still confused as to what is wrong with the limits. There is something wrong with it, but I don't think it is the reason you gave. When i took the limit, isn't $x$ a constant? – Lemon Jun 23 '13 at 23:04
  • It is, certainly, but in our definition, it is not, so there's potential for confusion. (Note also that we want the limit as $\epsilon\to 0^+$). To avoid that, we can rewrite our definition, and show that for all $c>0,$ there is some $d>0$ such that $$0<\epsilon-0<d\implies|(x+\epsilon)-x|<c,$$ or equivalently, $$0<\epsilon<d\implies|\epsilon|<c.$$ But this is simple. Take any $c>0,$ and put $d=c,$ so that for $0<\epsilon<d,$ we have $|\epsilon|=\epsilon<d=c.$ The approach works just fine. I merely wanted to advise caution. – Cameron Buie Jun 23 '13 at 23:17
  • Okay your example confuses me. I am failing to see what it has to do with my proof. Sorry; also do you mind cleaning up with comment box? Just easier on the both of us – Lemon Jun 23 '13 at 23:35
  • No problem. Since you deleted yours, it just looked like I was talking to myself, anyway. ;-) Now, which example do you mean? I wrote two proofs, but no examples (in our recent exchange here). – Cameron Buie Jun 23 '13 at 23:39
  • Your latest one with the $c$s and $d$s. I don't see what it has to do with our problem, at all. – Lemon Jun 23 '13 at 23:44
  • What if I changed my original problem as $y \leq x \iff y \leq x + z \forall z > 0$? And take $z \to 0$, still a problem? – Lemon Jun 23 '13 at 23:48
  • Well, if we hadn't replaced them, and just used the standard definition of limit as a template, then we'd have to show that $$\forall\epsilon>0,\exists\delta>0:0<\epsilon-0<\delta\implies|(x+\epsilon)-x|<\epsilon.$$ But this is impossible, which would suggest that $\lim_{\epsilon\to 0^+}x+\epsilon\ne x,$ if we didn't realize that we just used $\epsilon$ in two different contexts! – Cameron Buie Jun 23 '13 at 23:48
  • Changing the problem to $y\le x\iff y\le x+z:\forall z>0$ and taking $z\to 0^+$ actually works just fine to get us around the issue. (Note that that $z$ will not be the same as the one from my answer above.) In hindsight, there was really no need to replace the $\delta$.... – Cameron Buie Jun 23 '13 at 23:51
  • First we have $y \leq y + w \leq x + x$ and on the other hand we have $\lim_{w \to 0} y \leq \lim_{w \to 0} x + w = x$. $\blacksquare$ – Lemon Jun 24 '13 at 00:00
  • Excuse me, your above refers to your answer. I thought you were referring to your comment. I changed $z$ to $w$ – Lemon Jun 24 '13 at 00:03
  • That's just fine, if you've got the results that the limit of a constant is the constant, and that limits ignore added constants. Given your other post and your professed confusion of how to use $\epsilon$s in limit proofs, I thought you might be trying to prove the rigorously from the definition of limits, instead. Apologies for the sidetrack, but I'm glad to see that you've progressed! – Cameron Buie Jun 24 '13 at 00:04
  • But now I know why you think my proof was bad. You were saying conventionally we use $\epsilon$ in our definition, that's why you had all that $c$ and $d$. – Lemon Jun 24 '13 at 00:04
  • Yep! That's exactly what I was afraid of. – Cameron Buie Jun 24 '13 at 00:05
  • fine refers to my new proof right? I am not so sure about my $w \to 0$ part. I think it is wrong in some ways, that's why i asked in the beginning. – Lemon Jun 24 '13 at 00:06
  • and the reason why I felt it was wrong was because it seemed "sloppy" and I took the "shortcut" – Lemon Jun 24 '13 at 00:08
  • Your new proof is almost perfect (and will be perfect if you replace $w\to 0$ by $w\to 0^+$), just as long as you have the results to justify all the stages of $$y=\lim_{w\to 0^+}y\le\lim_{w\to 0^+}x+w=x.$$ Otherwise, you can still get an easy rigorous proof by showing that $$\forall\epsilon>0,\exists\delta>0:0<w<\delta\implies|(x+w)-x|<\epsilon.$$ Pretty much all we have to do there is take an arbitrary $\epsilon>0,$ then put $\delta=\epsilon,$ and the rest falls into place easily. – Cameron Buie Jun 24 '13 at 00:11
  • What's wrong with simply $w \to 0$? Your second mark on the $\forall \epsilon \dots$ is proving another problem now no? – Lemon Jun 24 '13 at 00:15
  • The problem is that "$w\to 0$" allows $w$ to be *negative* which we don't want to do, because then we don't get the $\le$ anymore, as we only made the assumption that $y\le x+w$ for all positive $w$. – Cameron Buie Jun 24 '13 at 00:27
  • Oh okay, basically means we have a very small negative in which case we realy don't get $<$, let alone $\leq$ is what you are sying? – Lemon Jun 24 '13 at 00:31
  • Pretty much. Since we only made an assumption for $w$ of positive sign, we need to be sure we don't accidentally let $w$ be negative. – Cameron Buie Jun 24 '13 at 00:35
  • Going back to the OP (me)'s original problem. Why didn't the author went for julien's solution right away instead of dealing with $\epsilon$s? His solution is more elementary. And why stop at $\epsilon > 0$? Let's make our lives easier by saying $2\epsilon >0$, so we can begin choosing $\sup A - x < \epsilon$ and similarly for $B$ where $x \in A$ – Lemon Jun 24 '13 at 00:37
  • I'm afraid I have no idea why the author went about it the way he did. I prefer julien's approach, myself (and upvoted his answer, despite the fact that it didn't actually answer your question). As for your bit with the $2\epsilon$, I'm not sure what you mean. Can you elaborate a bit? – Cameron Buie Jun 24 '13 at 00:54
  • I suppose it's possible that the author simply wanted to demonstrate a more generally applicable method of proving a non-strict inequality. – Cameron Buie Jun 24 '13 at 01:13
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Edit: I now realize this does not really answer the question, but that was already done several times. Instead, this is how I think you should prove this inequality. No $\epsilon$.

Fix $b\in B$. Then for every $a\in A$ $$ a=a+b-b\leq \sup (A+B)-b \quad\Rightarrow\quad \sup A\leq \sup (A+B)-b. $$ So for every $b\in B$ $$ b\leq \sup (A+B)-\sup A\quad\Rightarrow\quad \sup B\leq \sup (A+B)-\sup A. $$

Note: all this uses is the fact that the sup is the lub. So whenever you exhibit an upper bound, it is not smaller than the sup. Of course, you need to treat the case $A$ or $B$ unbounded above separately: $+\infty=+\infty$.

Julien
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  • Isn't it meangingless to talk about the supremum if the set isn't bounded in the first place? – Lemon May 09 '13 at 06:53
  • @sizz When $A$ is not bounded above, one defines $\sup A:=+\infty$. When it is not bounded below, one sets $\inf A:=-\infty$. That's very common. – Julien May 09 '13 at 12:16
  • But then what would be the point of defining the equality in this question? – Lemon May 09 '13 at 18:47
  • @sizz And why not, if it is true? It is just a limit case. And it says something, namely: $A+B$ is not bounded above if and only if $A$ or $B$ is not bounded above. – Julien May 09 '13 at 18:50
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$A$ and $B$ are sets:

For any $a\in A$ and $b\in B$, we have $a\le\sup(A)$ and $b\le\sup(B)$. Therefore, $$ a+b\le\sup(A)+\sup(B) $$ Thus, $$ \sup(A+B)\le\sup(A)+\sup(B)\tag{1} $$ $A+B=\{a+b:a\in A\text{ and }b\in B\}$. Thus, for any $a\in A$ and $b\in B$, $$ a+b\le\sup(A+B) $$ Taking the $\sup$ over $A$, we get $$ \sup(A)+b\le\sup(A+B) $$ then taking the $\sup$ over $B$, $$ \sup(A)+\sup(B)\le\sup(A+B)\tag{2} $$ Combining $(1)$ and $(2)$ yields $$ \sup(A+B)=\sup(A)+\sup(B) $$

$A$ and $B$ are functions

$$ \sup_{x\in X}(A(x)+B(x))=\sup_{\substack{x\in X\\y\in X\\x=y}}(A(x)+B(y))\le\sup_{\substack{x\in X\\y\in X}}(A(x)+B(y))=\sup_{x\in X}A(x)+\sup_{x\in X}B(x) $$ The inequality is because the $\sup$ is being taken over a smaller set on the left.

robjohn
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  • I initially solved it this way but am still not sure why we can just take the sup over A and B. Why are we allowed to do this and keep the same inequality? – Derek Luna Feb 10 '20 at 04:34
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If $x$ and $y$ are real numbers, then $x\leq y$ if (and only if) for all $\varepsilon>0$, $x\leq y+\varepsilon$.

To see this, it is easiest to use contraposition (and there is some repetition of what Cameron Buie posted as I write). Suppose that $x>y$. Then for the choice of $\varepsilon =\dfrac{x-y}{2}>0$, we have $x>y+\varepsilon$.

Jonas Meyer
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For all real numbers $s,t$, the assertion $s+\epsilon \ge t$ for all $\epsilon >0$ is equivalent to the assertion that $s\ge t$ (prove this fact if you are not sure why it is true). The claim you are asking about is a special case of this fact.

Ittay Weiss
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You should convince yourself that if $x\geq y$ for all $x\in X$ then $\inf X\geq y$.

Now, in the spirit of your question, consider

$X=\{\sup(A+B)+\epsilon\;|\;\epsilon>0\}$ and $y=\sup A+\sup B$. It should be immediate that $\inf X=\sup(A+B)$. Therefore, the claim follows.

mathemagician
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