Taken from sec. 1.4.1 of the book by Mary Hart, titled: Guide to Analysis.     Let $A, B$ be two non-empty sets of real numbers with supremums $\alpha, \beta$ respectively, and let the sets $A + B$ and $AB$ be defined by :   $A + B = {a + b: a\in A, b\in B}$,   $AB= {ab:a\in A, b\in B}$.   Show that $\alpha+ \beta$ is the supremum of A + B.   My attempt :  If a non-empty set of reals $A$ is having supermum $\alpha$, &amp; other set $B$ is the supremum $\beta$; then two cases are possible for each supermum:  (i) supremum is in set, and then it equals the maximum; as say $(0.1,0.2]$,  (ii) supremum is not in set, &amp; maximum does not exist; as say $(0.1, 0.2)$.   In case of combination of such two sets possibilities arise based on larger value. However, its inclusion, or exclusion (i.e. $],)$) does not affect supermum, as it always exists.  Say, set $A$ is $(a,b)$ &amp; set $B$ is $(c,d)$, then the sum has upper limit decided by the relative magnitudes of $a,b,c,d$. So, if $a\gt b \gt c \gt d$; then supermum is $b$, &amp; no maximum.   But, it is not clear how to prove the question.  P.S. There are few other questions from the same book, &amp; request link to their answers; as could not find (even, could not find answer to this!).  Till now, I resisted stating these 'other' questions, as wanted to post seperate questions (to prevent clutter), but if links were provided then no need. These questions are stated below:   Give an example to show that AB need not have a supremum. Prove also that even if AB has a supremum, this supremum need not be equal to $\alpha \beta$.  Show that if $A$ be set of positive reals with supremum $\alpha$, &amp; let $Y = {x^2 : x\in X}$; then $\alpha^2$ is supremum of Y.
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There are two sets $A$ and $B$ which are bound and are not empty. Now we'll define another set as: $A + B = \{a + b | a \in A, b \in B\}.$  I need to prove that $\sup(A + B) = \sup A + \sup B.$   I don't know if the way in which I proved it is enough or is missing something. This is my proof:  I know that for every $a \in A, b \in B$: $\sup A \geq a$ and $\sup B \geq b.$  Therefore $\sup A + sup B \geq a + b.$ I'll take a $c \in A+B$ and by $A+B$ definition: $c = a + b$. Then, it is safe to say that $c = a + b \leq \sup A + \sup B$ which gives: $c \leq \sup A + \sup B.$  I know that $c \leq \sup A + \sup B$ is true for every general $c \in A+B$, so the supremum of $A + B$ is: $\sup(A + B) = \sup A + \sup B.$  Is this a way to prove it or is it not enough? Thanks.
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A+B={a+b} I proved that the set A+B is bounded below. Now I'm stuck on how to prove that inf(A+B) = infA + infB
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Let $E$ be an empty set. Then, $\sup(E) = -\infty$ and $\inf(E)=+\infty$. I thought it is only meaningful to talk about $\inf(E)$ and $\sup(E)$ if $E$ is non-empty and bounded?   Thank you.
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Suppose $D$ is a nonempty bounded subset of reals. Let $f:D \to \mathbb R$ and $g:D \to \mathbb R$. Define $(f+g)(x)=f(x)+g(x)$. Prove $\sup(f+g)(D) \le \sup f(D) + \sup g(D)$ (also prove that $\sup (f+g)$ exists).   I understand why this is the case, just not how to prove it. Left side is pretty much $\sup  (f(x)+g(x))$ and right side is $\sup (f(x) + g(y))$ for $x,\,y \in D$. Basically $f+g$ has to use the same variable and $f(D)+g(D)$ use different ones. But I don't know how to go about proving this.     The second part of the question is to find a specific example where strict inequality holds.    Let $D=\{a,b\}$ and $f: a \to 1,\, b\to 0, \,g: a \to 0,\, b\to 1$. $\sup f(D) = 1,\, \sup g(D) = 1,\, \sup f(D) + \sup g(D) = 2.$ $\sup (f+g)(D) = 1$ (if we choose a, $f+g = 1+0,\, b,\, f+g=0+1$).
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In Zermelo-Fraenkel set theory do there exist sets $a,b$ such that $a \in b$ and $b \in a$. I think that no such sets exist but I am not sure how to prove why this is the case.
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I came to point where I suppose for case 1 that $A\subseteq B$ and conclusion is trivial. For case 2 I suppose that $A\not\subseteq B$ and try to prove $B\subseteq A$, but that gets me nowhere. Any pointers here are most welcome.
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A partially ordered set $(X, \leq)$ is called a lattice if for every pair of elements $x,y \in X$ both the infimum and suprememum of the set $\{x,y\}$ exists. I'm trying to get an intuition for how a partially ordered set can fail to be a lattice. In $\mathbb{R}$, for example, once two elements are selected the completeness of the real numbers guarantees the existence of both the infimum and supremum. Now, if we restrict our attention to a nondegenerate interval $(a,b)$ it is clear that no two points in $(a,b)$ have either a suprememum or infimum in $(a,b)$.   Is this the right way to think of a poset that is not a lattice? Is there perhaps a more fundamental example that would yield further clarity?
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Define $S\equiv\{ x\in \mathbb{Q}\mid x^2&lt;2\}$. Show that $\sup S=\sqrt{2} $.     For this question, I think that I would use the completeness axiom. As $3$ is greater than $2$, so $S$ has a upper bound. To yield a contradiction, I think I need to find there exists an constant $c$ which is a least upper bound such that $b^2&gt;2$, then there exists a number $r$ such that $b-r$ also in upper bound of $S$ which gives a contradiction since $b$ is the least upper bound of $S$. Then we can end the proof.    Does the idea right? If not, can anyone give a suggestion or a hit to write a better proof? Thanks.
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Edit: I was trying to find a general form of the question but I will just put the real question.  My textbook says |a-b|&lt;|b|/2 implies |a|>|b|/2. I don't see how and the text doesn't provide a proof.
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Claim. Let $a$ and $b$ be real numbers. If $a\geq b$  then $(a,b), [a,b)$ and $(a,b]$ are empty set.   My is question is that why?
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Claim: If $(x_\alpha)_{\alpha\in A}$ is a collection of real numbers $x_\alpha\in [0,\infty]$ such that $\sum_{\alpha\in A}x_\alpha&lt;\infty$, then $x_\alpha=0$ for all but at most countably many $\alpha\in A$ ($A$ need not be countable).  Proof: Let $\sum_{\alpha\in A}x_\alpha=M&lt;\infty$. Consider $S_n=\{\alpha\in A \mid x_\alpha&gt;1/n\}$.  Then $M\geq\sum_{\alpha\in S_n}x_\alpha&gt;\sum_{\alpha\in S_n}1/n=\frac{N}{n}$, where $N\in\mathbb{N}\cup\{\infty\}$ is the number of elements in $S_n$.  Thus $S_n$ has at most $Mn$ elements.   Hence $\{\alpha\in A \mid x_\alpha&gt;0\}=\bigcup_{n\in\mathbb{N}}S_n$ is countable as the countable union of finite sets. $\square$  First, is my proof correct? Second, are there more concise/elegant proofs?
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If $A,B$ non empty, upper bounded sets and $A+B=\{a+b\mid a\in A, b\in B\}$, how can I prove that $\sup(A+B)=\sup A+\sup B$?
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Let $A,B$ be any two sets. I really think that the statement $|A|\leq|B|$ or $|B|\leq|A|$ is true. Formally:     $$\forall A\forall B[\,|A|\leq|B| \lor\ |B|\leq|A|\,]$$   If this statement is true, what is the proof ?
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I read in a paper on set theory that the supremum and the infimum of the empty set are defined as $\sup(\{\})=-\infty$ and $\inf(\{\})=\infty$. But intuitively I can't figure out why that is the case. Is there a reason why the supremum and infimum of the empty set are defined this way?
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Let $A$ be a nonempty set of real numbers which is bounded below. Let $-A$ be the set of all numbers $-x$, where $x \in A$. Prove that $\inf A = -\sup(-A)$.   We know that $-A$ is bounded above. Hence $\sup(-A)$ exists. So to proceed, one would show that $\inf A \leq -\sup(-A)$ and $\inf A \geq -\sup(-A)$? Or $-\inf A &gt; \sup(-A)$ and $-\inf A &lt; \sup(-A)$.  I get the sense that one can use the following fact as well: (i) Let $\alpha = \sup S$. If $\beta &lt; \alpha$ then $\beta$ is not an upper bound for $S$.
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Now let set $A$ be arbitrary. So set $A$ exists. Using axiom of pairing $\{A\}$ exists. And then using axiom of regularity, we can argue that $A \notin A$. So, there is some $x$ such that $x \notin A$. Is this a valid proof ? Thanks
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Is my proof correct? Prove: $a_n \leq b_n \implies \limsup a_n \leq \limsup b_n$  Proof:  Let $a_n$ and $b_n$ be sequences such that $a_n \leq b_n \forall_n$. Suppose $\limsup a_n \nleq \limsup b_n$.   That is: $\limsup a_n &gt; \limsup b_n$. From this we know:  $\forall_{\epsilon &gt; 0} \exists_N \forall_{n&gt;N} \implies |b_n - b| &lt;\epsilon$. Where $b$ is the $\limsup b_n$.  $\forall_{\epsilon_1 &gt; 0} \exists_{N_1} \forall_{n &gt; N_1} \implies |a_n - a| &lt; \epsilon_1$. Where $a$ is the $\limsup a_n$  So, let $a^* = a + \dfrac{\epsilon_1}{2}$ and let $b^* = b - \dfrac{\epsilon}{2}$.  Hence, $a^* \in |a_n - a| &lt;\epsilon_1$ and $b^* \in |b_n - b| &lt; \epsilon$. And clearly we see that $b^* &lt; a^*$.Thus, we have found an element of $b_n$ namely $b^* &lt; a^*$ an element of $a$. This is contradiction since we are given $a_n \leq b_n \forall_n$.   Therefore, the supposition is false, and $\limsup a_n \leq \limsup b_n$.