I took $\sup A=\alpha$ and $\sup A=\beta$ I have to show that $\sup(A+B) = \sup A+\sup B$. I have showed that $\sup(A+B)$ exists first and also that $\alpha$ and $\beta$ is upper bound for set $A+B$. I just need to show that it is the least.

Claim: $\alpha + \beta$ is least upper bound

So I took $\gamma$ < $\alpha$ + $\beta$ where $\gamma$ is least upper bound. I am aiming to show that $\gamma=\alpha + \beta$

So I write $a_1+b_1 > \gamma - \epsilon$ for all epsilons, $a_1$ belongs to $A_1$ and so

$a_1+b_1 > \gamma - \epsilon$

Also $a_1 +b_1 \leq \alpha + \beta$

So i get $\gamma - \epsilon \leq a_1 +b_1 \leq \alpha + \beta $ for all e[silons

So $\gamma -(\alpha + \beta) < \epsilon$ for all $\epsilon >0$ and so I am done


Thanks for help

  • $\begingroup$ May i know why the downvotes? i have already shown my attempt along with question $\endgroup$ – Gathdi Aug 8 '16 at 3:54

Sorry, but the proof is incorrect. Let me rewrite it in a better way.

Let $\gamma$ be the lowest upper bound for $A+B$. By definition, if $\varepsilon>0$, there are $a\in A$ and $b\in B$ such that $$ a+b>\gamma-\varepsilon $$ On the other hand, $a+b\le\alpha+\beta$, so $$ \gamma-\varepsilon<\alpha+\beta $$ and therefore $$ \gamma-(\alpha+\beta)<\varepsilon $$ Since $\varepsilon$ is arbitrary, you conclude that $\gamma-(\alpha+\beta)=0$, but this is wrong: all you can say is that $$ \gamma-(\alpha+\beta)\le0 $$ which is already known, as clearly $\alpha+\beta$ is an upper bound for $A+B$.

What's the problem? That you never used the assumptions that $\alpha=\sup A$ and $\beta=\sup B$.

You can directly prove instead that $\alpha+\beta$ is the lowest upper bound. It clearly is an upper bound.

Let $\varepsilon>0$. Since $\alpha=\sup A$, there is $a\in A$ with $a>\alpha-\frac{\varepsilon}{2}$. Since $\beta=\sup B$, there is $b\in B$ with $b>\beta-\frac{\varepsilon}{2}$. Then $$ a+b>\left(\alpha-\frac{\varepsilon}{2}\right)+ \left(\beta-\frac{\varepsilon}{2}\right)=(\alpha+\beta)-\varepsilon $$

  • $\begingroup$ If $a<\epsilon$ for all epsilon>0. Is it not that a=0 $\endgroup$ – Gathdi Aug 2 '16 at 10:21
  • $\begingroup$ $a+b \leq \alpha + \beta$. This is where i used assumption $\endgroup$ – Gathdi Aug 2 '16 at 10:23
  • $\begingroup$ @Gathdi First comment: no, if $a<\varepsilon$ for all $\varepsilon>0$, then $a\le0$; you can conclude that $a=0$ only if you already that $a\ge0$. Second comment: you're only using that $\alpha$ and $\beta$ are upper bounds for $A$ and $B$, not that they're the lowest upper bounds. $\endgroup$ – egreg Aug 2 '16 at 10:25
  • $\begingroup$ I am unsure as to how what you have done have proved result i aimed at $\endgroup$ – Gathdi Aug 2 '16 at 10:31
  • $\begingroup$ @Gathdi I proved that $\alpha+\beta$ satisfies both properties for the lowest upper bound. Recall that $\gamma$ is the lowest upper bound for $C$ if (1) it is an upper bound for $C$ (that is, for all $c\in C$, $c\le\gamma$); (2) if $\varepsilon>0$, there is $c\in C$ such that $c>\gamma-\varepsilon$. $\endgroup$ – egreg Aug 2 '16 at 10:41

I think what you write is fine, nevertheless I will give it a go.

Now, we have the sets A and B, whose suprema are $\alpha$ and $\beta$ respectively. We are trying to show that the set $A+B = \{ x_1 + x_2, x_1 \in A ,x_2 \in B \}$ has supremum as $\alpha + \beta$.

On one hand, $x \leq \alpha \quad \forall x \in A$, and $y \leq \beta \quad \forall y \in B$, hence $x+y \leq \alpha + \beta \forall x \in A \forall y \in B$, so it follows that $A+B$ is upper bounded by $\alpha + \beta$.

Now, to show that $\alpha+\beta$ is the least upper bound, suppose there existed a strictly lower upper bound, say $L$. Then let $\epsilon = \alpha+\beta - L$. Clearly, $\epsilon > 0$.

Because $\alpha$ is the supremum of $A$, it follows that there is some $a \in A$ such that $\alpha - a < 0.5\epsilon$.

Similarly, because $\beta$ is the supremum of $B$, it follows that there is some $b \in B$ such that $\beta- b < 0.5\epsilon$.

Now, note that by adding the above, we get that $\alpha + \beta - (a+b) < \epsilon$, hence $(a+b)>L$. This is a contradiction because we assumed $L$ was an upper bound. Hence $\alpha + \beta$ is the supremum, and we are done.

Always know how to operate with epsilon-deltas. It will help in dealing with continuity etc. in the future.

  • $\begingroup$ Should it not in last paragarph that Alpha + Beta > L instead of a + b $\endgroup$ – Gathdi Aug 2 '16 at 10:36
  • $\begingroup$ No, what I have written is correct. Remember that because we are trying to show the opposite of "$L$ is an upper bound", we have to find two elements whose sum is greater than $L$. Note that $a$ is an element of $A$ and $b$ is an element of $B$. You can then see that $a+b$ is an element of $A+B$ which is greater than $L$ (as I showed above). That is the contradiction. We already assumed that Alpha + Beta > L, so writing it twice would not help. $\endgroup$ – астон вілла олоф мэллбэрг Aug 4 '16 at 3:13
  • $\begingroup$ @ астон вілла олоф мэллбэрг: Thank you for this solution! One vote for this! $\endgroup$ – Omojola Micheal Apr 11 '18 at 19:49
  • $\begingroup$ @Mike Thank you for your appreciation. $\endgroup$ – астон вілла олоф мэллбэрг Apr 11 '18 at 22:53

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