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I am self-studying through Amann & Escher, Analysis, Vol 1, Chapter 1. I created the following proof and wonder if I have succeeded. If I have (I feel that I have with the caveat that I wonder if I made too big of a leap possibly on one of the steps - I am so new at this that I don't have a sense of correctness or style yet firmly developed), then I'll give an answer credit to anyone who helps me improve the proof or otherwise learn something to make it more robust. If it is not yet a correct proof, then I'll award an answer to the individual who can point out why and a hint as to what I must do to make it satisfactory.

7) Let $X$ and $A$ be subsets of a set $U$ and let $Y$ and $B$ be subsets of a set V. Prove the following:

a) If $A\times B\ne\emptyset$, then $A\times B\subseteq X\times Y\Longleftrightarrow(A\subseteq X)\land(B\subseteq Y).$

We must show that $A\times B\subseteq X\times Y\Longrightarrow(A\subseteq X)\wedge(B\subseteq Y)$ and that $(A\subseteq X)\land(B\subseteq Y)\Longrightarrow A\times B\subseteq X\times Y.$

Proof.

``$\Longrightarrow$'':

This part of the proof is done by contradiction. Suppose $A\times B\subseteq X\times Y$ and $\lnot[(A\subseteq X)\land(B\subseteq Y)]$. $A\times B\subseteq X\times Y$ means that for all $a$ and for all $b$, if $(a,b)\in A\times B$ then $(a,b)\in X\times Y$. $\lnot[(A\subseteq X)\land(B\subseteq Y)]$ means that for some $a\in A,a\notin X$ or for some $b\in B,b\notin Y$. From this last sentence we deduce that there exists some pair $(a,b)\in A\times B$ such that $(a,b)\notin X\times Y$. But that contradicts our premise: $A\times B\subseteq X\times Y.$ Therefore, $\lnot[(A\subseteq X)\land(B\subseteq Y)]$ must be false and $(A\subseteq X)\land(B\subseteq Y)$ must be true. Hence, we have shown that $A\times B\subseteq X\times Y\Longrightarrow(A\subseteq X)\land(B\subseteq Y).$

``$\Longleftarrow$'':

This part of the proof is done by contradiction. Suppose $(A\subseteq X)\land(B\subseteq Y)$ and that the statement $A\times B\subseteq X\times Y$ is false. $(A\times B\not\subseteq X\times Y)$ means there is some $(a,b)\in A\times B$ such that $(a,b)\notin X\times Y$. From this we deduce that for some $a\in A,a\notin X$ and for some $b\in B,b\notin Y$. But that contradicts our premise: $(A\subseteq X)\land(B\subseteq Y)$. Therefore, $\lnot(A\times B\subseteq X\times Y)$ must be false and $A\times B\subseteq X\times Y$ must be true. Hence, we have shown that $(A\subseteq X)\land(B\subseteq Y)\Longrightarrow A\times B\subseteq X\times Y$. Thus, the statement ``if $A\times B\ne\emptyset$, then $A\times B\subseteq X\times Y\Longleftrightarrow(A\subseteq X)\land(B\subseteq Y)$'' must be true.

Joe
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    When you say "$\neg [(A \subseteq X) \land (B \subseteq Y)]$ means that for all $x \in A$, $a \not \in X$ and/or for all $b \in B$, $b \not \in Y$", this isn't true. The negation of the statement $p \land q$ is $\neg p \lor \neg q$. Also, the negation of $A \subseteq X$ is $\exists a \in A$ such that $a \not \in X$. It isn't necessarily true that for every $a \in A$, $a \not \in X$. It just has to be true for some $a \in A$. – layman Nov 13 '14 at 03:28
  • Thanks @MathIsHardNoItsNot. I've incorporated the items you caught/corrected into the edited version. – Joe Nov 13 '14 at 03:48
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    You're welcome. I said it earlier, but I think because I was saying multiple things at once, you missed it. The negation of the statement "$p$ and $q$" is "not $p$ or not $q$". So, when you write "and/or" when describing your negation, you should really just have "or". The "and" should not be there. – layman Nov 13 '14 at 03:48
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    If you made that change, too, then your "$\implies$" direction looks great. For the "$\impliedby$" direction, once you negate the statement $A \times B \subseteq X \times Y$, you say "[the negation] means that for some $a$ and $b$, if $(a,b) \in A \times B$, then $(a, b) \not \in X \times Y$". You have the right idea, but the language is wrong. If we know $A \times B \not \subseteq X \times Y$, then that doesn't mean the statement "if $(a,b) \in A \times B$, then $(a,b) \not \in X \times Y$" is true. $A \times B \not \subseteq X \times Y$ means there is some $(a,b) \in A \times B$ such tht – layman Nov 13 '14 at 03:49
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    $(a,b) \not \in X \times Y$. This is different than saying "if $(a, b) \in A \times B$, then $(a,b) \not \in X \times Y$". Does that make sense? – layman Nov 13 '14 at 03:52
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    By the way, my last comments were basically all of the changes I thought should be made. If those changes are made, then I think both directions look fine. – layman Nov 13 '14 at 03:56
  • Thanks for that last tip @MathIsHardNoItsNot. I incorporated the last suggestion you made for the second half of the proof. – Joe Nov 13 '14 at 04:56

2 Answers2

3

  1. $\left (A \subseteq X\right) \land \left (B \subseteq Y\right)\implies \left(A\times B\right) \subseteq \left(X\times Y\right)$

  2. $\left(A\times B\right) \subseteq \left(X\times Y\right) \implies (a,b)\in \left(X\times Y\right) \left(a\in A \land b \in B\right) \implies a\in X \land b\in Y \implies A\subseteq X \land B \subseteq Y$

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    So, really the OP was asking for a verification of his proof. – layman Nov 13 '14 at 03:53
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    @MathIsHardNoItsNot: That should be included in the tags. –  Nov 13 '14 at 03:55
  • Ok, I agree, but did you read his post? It seems like you just read the title and answered. – layman Nov 13 '14 at 03:56
  • @MathIsHardNoItsNot: Yes, but it seemed to be that the OP is using too much explanation. That's why I posted this answer to increase the compactness of the proof. –  Nov 13 '14 at 03:59
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    Oh, ok. I think at his level, he needs these details. He is essentially writing down his thought process to make sure he is thinking about it right. Would you write a proof that way in practice? Probably not, but we all start out there in the beginning. I understand the point of your answer, though. – layman Nov 13 '14 at 04:01
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    Thanks user170039 and MathIsHardNoItsNot. Couple interesting points for people coming later: the proof style is mimicking that provided in a similar proof on p.11 of Amann & Escher (referenced above). Also, while I can see the potential confusion of "and/or" for "or", I actually picked that up from Kolmogorov & Fomin, Vol. 1. Presumably, they use it to make clear to beginners that the or is not exclusive. It is unusual though, I realize (but I do sure like Kolmogorov's proofs!). Anyways, you guys are great for helping me. – Joe Nov 13 '14 at 05:02
  • @user170039, compact indeed! Thanks. – Joe Nov 13 '14 at 05:25
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$ \newcommand{\calc}{\begin{align} \quad &} \newcommand{\calcop}[2]{\\ #1 \quad & \quad \unicode{x201c}\text{#2}\unicode{x201d} \\ \quad & } \newcommand{\endcalc}{\end{align}} \newcommand{\ref}[1]{\text{(#1)}} $Just for fun, here is an alternative style of proof: start at the most complex side, expand the definitions, apply the laws of logic, simplify, and see where that leads.

It looks like the left hand side is the most complex one, and therefore we calculate as follows:

$$\calc A \times B \;\subseteq\; X \times Y \calcop\equiv{definition of $\;\subseteq\;$; definition of $\;\times\;$, twice} \langle \forall a,b : a \in A \land b \in B : a \in X \land b \in Y \rangle \calcop\equiv{logic: $\;\forall\;$ distributes over $\;\land\;$ -- there isn't much else we can do} \langle \forall a,b : a \in A \land b \in B : a \in X \rangle \;\land\; \langle \forall a,b : a \in A \land b \in B : b \in Y \rangle \calcop\equiv{logic: separate the quantifications, twice -- to separate $\;a\;$ from $\;b\;$} \langle \forall b : b \in B : \langle \forall a : a \in A : a \in X \rangle \rangle \;\land\; \langle \forall a : a \in A : \langle \forall b : b \in B : b \in Y \rangle \rangle \calcop\equiv{logic: write $\;P \Rightarrow Q\;$ as $\;\lnot P \lor Q\;$ and extract $\;{} \lor Q\;$ out of $\;\forall\;$, twice} (\langle \forall b :: b \not\in B \rangle \lor \langle \forall a : a \in A : a \in X \rangle) \;\land\; (\langle \forall a :: a \not\in A \rangle \lor \langle \forall b : b \in B : b \in Y \rangle) \calcop\equiv{definitions of $\;\emptyset\;$ and $\;\subseteq\;$, both twice} (B = \emptyset \lor A \subseteq X) \;\land\; (A = \emptyset \lor B \subseteq Y) \endcalc$$

And if we're given that $\;A,B\;$ are both non-empty, then that last line simplifies to $\;A \subseteq X \;\land\; B \subseteq Y\;$, so that we've proven $$ A \times B \;\subseteq\; X \times Y \;\equiv\; A \subseteq X \;\land\; B \subseteq Y $$ as required.

MarnixKlooster ReinstateMonica
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  • Thanks Marnix. I have a couple questions: (1) I haven't seen the outer angle brackets used in this context before except for Cartesian products. You seem to know a notation that I haven't been exposed to which puts quantifiers inside angle brackets. I've seen Patrick Suppes notation for quantifiers in his Axiomatic Set Theory book. Can you recommend a reference that contains such notation? I am building a library. (2) All the steps are clear to me except step 3 and 4. I understand them but I haven't seen that sort of set factoring sequence before or the compound such that notation. – Joe Nov 20 '14 at 01:26
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    @Joe For your question (1): I've learned this notation from Dijkstra et al.; see the end of my answer http://math.stackexchange.com/a/332186/11994 for several references. Specifically, the brackets notation and its properties is discussed in section "Quantification" of EWD1300. – MarnixKlooster ReinstateMonica Nov 20 '14 at 20:49
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    @Joe On (2): Step (3) uses the rule, which hold for many quantifiers $;\odot;$, like $;\forall,\exists,\sum;$, etc., that $$\langle \odot x :: \langle \odot y : P(x) \land P(y) : Q(x,y) \rangle \rangle ;=; \langle \odot x : P(x) : \langle \odot y : P(y) : Q(x,y) \rangle \rangle$$ since $;P(x);$ does not contain $;y;$. In the specific case of $;\forall;$ the key step is to rewrite $;\langle \forall y :: \lnot P(x) \lor \lnot P(y) \lor Q(x,y) \rangle;$ to $;\lnot P(x) \lor \langle \forall y :: \lnot P(y) \lor Q(x,y) \rangle;$. Step (4) uses a very similar key step. – MarnixKlooster ReinstateMonica Nov 20 '14 at 21:03
  • Thanks Marnix for answering my questions and for the reference request answer. – Joe Dec 08 '14 at 19:49