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1d
comment How has the teaching of (undergraduate) Set Theory changed over time?
You might look at when real analysis textbooks started including the now obligatory introductory chapter on set-theoretic preliminaries. I'm pretty sure they wouldn't have in Gauss's day. And what aspects of set theory they covered and how that may have changed over the years and why. Let us know what you find
2d
comment Why can we assume a statement is true for $n = k$, when using induction?
Imagine you are holding a bucket of natural numbers. Suppose you can show that the number 1 is in your bucket. Suppose further that if you find any number (k) in your bucket, you will somehow be able to find the next number (k+1) in there. Since 1 is in there, 2 must also be in there. Since 2 is there, 3 must also be in there. And so on. It stands to reason that you must have ALL natural numbers in your bucket
2d
comment Example of a collection that is not a set?
Collections of objects that are not sets don't seem to come up very much if at all in mainstream mathematics. Yes, I'm sure there are rarefied domains in which they may be indispensable, but you can probably do all of classical analysis without them.
Apr
26
revised If $G$ is a finite group and $H \subset G$ is closed, must $H$ be a subgroup?
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Apr
25
asked If $G$ is a finite group and $H \subset G$ is closed, must $H$ be a subgroup?
Apr
21
comment How to prove that a set exists in ZFC?
You can only prove that a set exists if you can construct if from sets that are assumed to exist in your set theory. If your set theory assumes the existence only the empty set, for example, you can only prove the existence of sets that are constructed starting with the empty set and using the other axioms. You could, of course, simply begin a proof with, say, some version of Peano's axioms and start deriving the theorems of number theory. If these axioms turn out to be inconsistent (highly unlikely), you would have to modify or abandon them altogether.
Apr
21
comment Is PA the most common foundation for arithmetic?
The term "Peano Arithmetic" (PA) has been more or less appropriated by the first-order crowd who disavow set theory and define addition and multiplication in their version of Peano's Axioms. The most common foundation for arithmetic, however, would seem to be Peano's Axioms whittled down to only the five axioms that define a successor function on the set of natural numbers, together with rules and axioms of logic and set theory. Addition, multiplication and exponentiation on the natural numbers can be constructed from these five axioms using the rules and axioms of logic and set theory.
Apr
20
revised Prove that: $\vdash \forall x \exists !y(y=x)$
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Apr
20
revised Prove that: $\vdash \forall x \exists !y(y=x)$
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Apr
20
revised Prove that: $\vdash \forall x \exists !y(y=x)$
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Apr
20
answered Prove that: $\vdash \forall x \exists !y(y=x)$
Apr
19
revised Given an arbitrary truth table, is it always possible to construct a sequent that satisfies it?
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Apr
19
answered Given an arbitrary truth table, is it always possible to construct a sequent that satisfies it?
Apr
18
comment Peano and induction according to Schaum's Outline
Using the axioms of set theory, you can construct the addition functions starting with the 5 Peano Axioms, i.e. you can prove the existence of an add function $+$ such that $x+0=x$ and $x+s(y)=s(x+y)$. Shaum's no doubt skips over this point, but there is nothing "hand wavy" about Peano's axioms.
Apr
15
comment What is the meaning of the notation $A :\Leftrightarrow B$?
That is equivalent, in words, to: B is true whenever A is true.
Apr
14
comment What is the meaning of the notation $A :\Leftrightarrow B$?
How does the author define $\implies$?
Apr
14
comment Why include equality in FOL for ZFC?
How else do you determine if two sets are equal (the same) if not by looking at their respective elements?
Apr
13
comment What is the meaning of the notation $A :\Leftrightarrow B$?
Could be a misprint. I would add: B is true whenever A is true. As it stands, it defines $B\implies A$.
Apr
13
comment How can the law of the excluded middle possibly be true if we acknowledge that some logical statements are undefined?
Re: 1/0 example. We define division on $R$ as follows: $\forall a,b,c\in R:[b\neq 0 \implies [a/b=c \iff a=cb]]$. Is $1/0=0$? We cannot use this definition to determine if it is true or not. We cannot apply the definition in this case because we have $0$ denominator ($b=0$). In general, we say that any expression with a $0$ denominator is undefined. Nothing mysterious about that. LEM still applies. Being "undefined" is not some kind of third value for logical expressions. It just means that our definition does not handle that case.
Apr
11
revised Peano and induction according to Schaum's Outline
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