Subclass of ordinals is a set iff it is boundedThis set of ordinals is $in$-transitive?Euclidean algorithm for ordinalsProve using transfinite induction that if ordinals $alpha$ and $beta$ are countable, then so is $alpha + beta$.the smallest transfinite ordinal numberProve that elements of ordinals are ordinals.Do I need that $mathrmOrd$ is well-founded before proving that it is totally ordered?(Verification) well-definiteness of order between ordinalsProving that the class of all ordinals is well ordered under $in$isomorphism from well ordered set to ordinalLet $alpha,gamma$ be ordinals such that $0<alphalegamma$. Then there is a greatest ordinal $beta$ such that $alphacdotbetalegamma$
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Subclass of ordinals is a set iff it is bounded
This set of ordinals is $in$-transitive?Euclidean algorithm for ordinalsProve using transfinite induction that if ordinals $alpha$ and $beta$ are countable, then so is $alpha + beta$.the smallest transfinite ordinal numberProve that elements of ordinals are ordinals.Do I need that $mathrmOrd$ is well-founded before proving that it is totally ordered?(Verification) well-definiteness of order between ordinalsProving that the class of all ordinals is well ordered under $in$isomorphism from well ordered set to ordinalLet $alpha,gamma$ be ordinals such that $0<alphalegamma$. Then there is a greatest ordinal $beta$ such that $alphacdotbetalegamma$
$begingroup$
Let $Xsubseteq textOrd$. Then $X$ is a set if and only if there exists an ordinal $beta$ such that for all $alpha in X$, $betage alpha$.
I am really having trouble with proving that something is a set or using the fact that we have a set. How does one approach problems like this, when you have to prove something is a set?
elementary-set-theory ordinals
asked Mar 13 at 18:58
Analysis801Analysis801
1317
$endgroup$
add a comment |
$begingroup$
Let $Xsubseteq textOrd$. Then $X$ is a set if and only if there exists an ordinal $beta$ such that for all $alpha in X$, $betage alpha$.
I am really having trouble with proving that something is a set or using the fact that we have a set. How does one approach problems like this, when you have to prove something is a set?
elementary-set-theory ordinals
asked Mar 13 at 18:58
Analysis801Analysis801
1317
$endgroup$
add a comment |
$begingroup$
Let $Xsubseteq textOrd$. Then $X$ is a set if and only if there exists an ordinal $beta$ such that for all $alpha in X$, $betage alpha$.
I am really having trouble with proving that something is a set or using the fact that we have a set. How does one approach problems like this, when you have to prove something is a set?
elementary-set-theory ordinals
asked Mar 13 at 18:58
Analysis801Analysis801
1317
$endgroup$
Let $Xsubseteq textOrd$. Then $X$ is a set if and only if there exists an ordinal $beta$ such that for all $alpha in X$, $betage alpha$.
I am really having trouble with proving that something is a set or using the fact that we have a set. How does one approach problems like this, when you have to prove something is a set?
elementary-set-theory ordinals
elementary-set-theory ordinals
asked Mar 13 at 18:58
Analysis801Analysis801
1317
asked Mar 13 at 18:58
Analysis801Analysis801
1317
asked Mar 13 at 18:58
Analysis801Analysis801
1317
asked Mar 13 at 18:58
Analysis801Analysis801
1317
asked Mar 13 at 18:58
Analysis801Analysis801
1317
1317
add a comment |
add a comment |
1 Answer
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$begingroup$
It makes sense to take the axiom system you're working with (ZF I suppose?) and go from there. "x is a set" is just short for "ZF proves the existence of x". Depending on your formulation of the ZF axioms, one would approach this as follows: for example, to show that for sets $x,y$ the Kuratowski pair $langle x,y rangle = x , x,y $ is a set, we apply pairing twice to get a set $z$ that contains $langle x,y rangle$, and then use comprehension on $z$ to obtain the Kuratowski pair itself. Note we have assumed the existence of arbitrary sets and only used the axioms from there to derive the conclusion.
In order to show that some collection is in fact a proper class and hence not a set, a proof by contradiction is usually the way to go. In your particular example, indeed proceed by contradiction. If $C$ is an unbounded class of ordinals and we assume it is a set, is there an axiom of ZF that proves the existence of a set containing each ordinal in $C$? What is this set equal to?
answered Mar 13 at 19:41
MacRanceMacRance
1226
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$begingroup$
It makes sense to take the axiom system you're working with (ZF I suppose?) and go from there. "x is a set" is just short for "ZF proves the existence of x". Depending on your formulation of the ZF axioms, one would approach this as follows: for example, to show that for sets $x,y$ the Kuratowski pair $langle x,y rangle = x , x,y $ is a set, we apply pairing twice to get a set $z$ that contains $langle x,y rangle$, and then use comprehension on $z$ to obtain the Kuratowski pair itself. Note we have assumed the existence of arbitrary sets and only used the axioms from there to derive the conclusion.
In order to show that some collection is in fact a proper class and hence not a set, a proof by contradiction is usually the way to go. In your particular example, indeed proceed by contradiction. If $C$ is an unbounded class of ordinals and we assume it is a set, is there an axiom of ZF that proves the existence of a set containing each ordinal in $C$? What is this set equal to?
answered Mar 13 at 19:41
MacRanceMacRance
1226
$endgroup$
add a comment |
$begingroup$
It makes sense to take the axiom system you're working with (ZF I suppose?) and go from there. "x is a set" is just short for "ZF proves the existence of x". Depending on your formulation of the ZF axioms, one would approach this as follows: for example, to show that for sets $x,y$ the Kuratowski pair $langle x,y rangle = x , x,y $ is a set, we apply pairing twice to get a set $z$ that contains $langle x,y rangle$, and then use comprehension on $z$ to obtain the Kuratowski pair itself. Note we have assumed the existence of arbitrary sets and only used the axioms from there to derive the conclusion.
In order to show that some collection is in fact a proper class and hence not a set, a proof by contradiction is usually the way to go. In your particular example, indeed proceed by contradiction. If $C$ is an unbounded class of ordinals and we assume it is a set, is there an axiom of ZF that proves the existence of a set containing each ordinal in $C$? What is this set equal to?
answered Mar 13 at 19:41
MacRanceMacRance
1226
$endgroup$
add a comment |
$begingroup$
It makes sense to take the axiom system you're working with (ZF I suppose?) and go from there. "x is a set" is just short for "ZF proves the existence of x". Depending on your formulation of the ZF axioms, one would approach this as follows: for example, to show that for sets $x,y$ the Kuratowski pair $langle x,y rangle = x , x,y $ is a set, we apply pairing twice to get a set $z$ that contains $langle x,y rangle$, and then use comprehension on $z$ to obtain the Kuratowski pair itself. Note we have assumed the existence of arbitrary sets and only used the axioms from there to derive the conclusion.
In order to show that some collection is in fact a proper class and hence not a set, a proof by contradiction is usually the way to go. In your particular example, indeed proceed by contradiction. If $C$ is an unbounded class of ordinals and we assume it is a set, is there an axiom of ZF that proves the existence of a set containing each ordinal in $C$? What is this set equal to?
answered Mar 13 at 19:41
MacRanceMacRance
1226
$endgroup$
It makes sense to take the axiom system you're working with (ZF I suppose?) and go from there. "x is a set" is just short for "ZF proves the existence of x". Depending on your formulation of the ZF axioms, one would approach this as follows: for example, to show that for sets $x,y$ the Kuratowski pair $langle x,y rangle = x , x,y $ is a set, we apply pairing twice to get a set $z$ that contains $langle x,y rangle$, and then use comprehension on $z$ to obtain the Kuratowski pair itself. Note we have assumed the existence of arbitrary sets and only used the axioms from there to derive the conclusion.
In order to show that some collection is in fact a proper class and hence not a set, a proof by contradiction is usually the way to go. In your particular example, indeed proceed by contradiction. If $C$ is an unbounded class of ordinals and we assume it is a set, is there an axiom of ZF that proves the existence of a set containing each ordinal in $C$? What is this set equal to?
answered Mar 13 at 19:41
MacRanceMacRance
1226
answered Mar 13 at 19:41
MacRanceMacRance
1226
answered Mar 13 at 19:41
MacRanceMacRance
1226
answered Mar 13 at 19:41
MacRanceMacRance
1226
1226
add a comment |
add a comment |
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