Mixing
Clarification on set definitions from How to Prove it (Velleman) Chapter 2.3
$begingroup$
I am reading the first two paragraphs of Chapter 2.3 in How to Prove It (Velleman) and I am unclear about why $x in {n^2 | n in Bbb{N}}$ is the same as $exists n in Bbb{N}(x = n^2)$
I will lay out Velleman's explanation and my understanding of it, I would love some feedback on if I'm thinking about this correctly.
For the reader, up until this point in the book, we know two ways to define sets.
- list elements in brackets $${1, 2, 3, 4}$$
- use elementhood notation $${x | P(x)}$$
Velleman suggests that it is common to replace the $x$ in front of the vertical line with a more complex expression. He uses the example of $S$, the set of perfect squares
$$S = {n^2 | n in Bbb{N}}tag{1}$$
He claims that (1) can be written as
$$S = {x | exists n in Bbb{N}(x=n^2)}tag{2}$$
Thus
$$S = {n^2 | n in Bbb{N}} = {x | exists n in Bbb{N}(x=n^2)}$$
At this point I'm still with Velleman. Replacing x before the vertical line with a more complex expression is intuitive. Putting the statement "x is a perfect square" into the form $exists n in Bbb{N}(x=n^2)$ and defining the set S using that logical statement as an elementhood test - $S = {x | exists n in Bbb{N}(x=n^2)}$ - makes sense (this question is a good discussion. on eq 2)
Velleman continues
and therefore $x in {n^2 | n in Bbb{N}}$ means the same thing as $exists n in Bbb{N}(x = n^2)$
I am missing the jump to this final conclusion, here are my thoughts so far.
First thought
Perhaps a step Velleman is taking (and assumes reader will make) is that $x in {n^2 | n in Bbb{N}}$ is the same as $x in {x | exists n in Bbb{N}(x = n^2)}$. And then, using the following from Chapter 1.3,
in general, the statement $y in {x | P(x)}$ means the same thing as P(y)...a statement about y but not x.
from this, we can see that $x in {x | exists n in Bbb{N}(x = n^2)}$ means the same thing as $exists n in Bbb{N}(x = n^2)$, and then arrive at the conclusion.
Second thought (not entirely unrelated way of thinking about it)
$exists n in Bbb{N}(x = n^2)$ and $x in {n^2 | n in Bbb{N}}$ are both statements about x, where x is a free variable. These statement can be evaluated to true or false, but that will depend on the value of x that is used. They will always evaluate to the same truth/false value, therefore they mean the same thing.
elementary-set-theory logic proof-explanation predicate-logic
asked Jan 9 at 15:16
Jake KirschJake Kirsch
547
$endgroup$
|
show 2 more comments
$begingroup$
I am reading the first two paragraphs of Chapter 2.3 in How to Prove It (Velleman) and I am unclear about why $x in {n^2 | n in Bbb{N}}$ is the same as $exists n in Bbb{N}(x = n^2)$
I will lay out Velleman's explanation and my understanding of it, I would love some feedback on if I'm thinking about this correctly.
For the reader, up until this point in the book, we know two ways to define sets.
- list elements in brackets $${1, 2, 3, 4}$$
- use elementhood notation $${x | P(x)}$$
Velleman suggests that it is common to replace the $x$ in front of the vertical line with a more complex expression. He uses the example of $S$, the set of perfect squares
$$S = {n^2 | n in Bbb{N}}tag{1}$$
He claims that (1) can be written as
$$S = {x | exists n in Bbb{N}(x=n^2)}tag{2}$$
Thus
$$S = {n^2 | n in Bbb{N}} = {x | exists n in Bbb{N}(x=n^2)}$$
At this point I'm still with Velleman. Replacing x before the vertical line with a more complex expression is intuitive. Putting the statement "x is a perfect square" into the form $exists n in Bbb{N}(x=n^2)$ and defining the set S using that logical statement as an elementhood test - $S = {x | exists n in Bbb{N}(x=n^2)}$ - makes sense (this question is a good discussion. on eq 2)
Velleman continues
and therefore $x in {n^2 | n in Bbb{N}}$ means the same thing as $exists n in Bbb{N}(x = n^2)$
I am missing the jump to this final conclusion, here are my thoughts so far.
First thought
Perhaps a step Velleman is taking (and assumes reader will make) is that $x in {n^2 | n in Bbb{N}}$ is the same as $x in {x | exists n in Bbb{N}(x = n^2)}$. And then, using the following from Chapter 1.3,
in general, the statement $y in {x | P(x)}$ means the same thing as P(y)...a statement about y but not x.
from this, we can see that $x in {x | exists n in Bbb{N}(x = n^2)}$ means the same thing as $exists n in Bbb{N}(x = n^2)$, and then arrive at the conclusion.
Second thought (not entirely unrelated way of thinking about it)
$exists n in Bbb{N}(x = n^2)$ and $x in {n^2 | n in Bbb{N}}$ are both statements about x, where x is a free variable. These statement can be evaluated to true or false, but that will depend on the value of x that is used. They will always evaluate to the same truth/false value, therefore they mean the same thing.
elementary-set-theory logic proof-explanation predicate-logic
asked Jan 9 at 15:16
Jake KirschJake Kirsch
547
$endgroup$
$begingroup$
Surely (2) is simply Velleman's definition of what (1) means? Then it all makes sense, right? I don't have Velleman's book to hand, so I can't check this for myself.
$endgroup$
– TonyK
Jan 9 at 15:26
$begingroup$
I would say your first and second thought are both correct, and are in fact the same thought stated in different words.
$endgroup$
– Mees de Vries
Jan 9 at 15:27
$begingroup$
@TonyK it does make sense and I agree they are equal, but I am very new to this topic and I haven't studied math in a long time! I am trying to understand exactly why the definitions are the same, based on the concepts presented in the text thus far.
$endgroup$
– Jake Kirsch
Jan 9 at 15:31
$begingroup$
You have completely misunderstood my comment! Please read it more carefully.
$endgroup$
– TonyK
Jan 9 at 15:34
$begingroup$
We have that $x in { z mid P(z) } text { iff } P(x)$.
$endgroup$
– Mauro ALLEGRANZA
Jan 9 at 15:48
|
show 2 more comments
$begingroup$
I am reading the first two paragraphs of Chapter 2.3 in How to Prove It (Velleman) and I am unclear about why $x in {n^2 | n in Bbb{N}}$ is the same as $exists n in Bbb{N}(x = n^2)$
I will lay out Velleman's explanation and my understanding of it, I would love some feedback on if I'm thinking about this correctly.
For the reader, up until this point in the book, we know two ways to define sets.
- list elements in brackets $${1, 2, 3, 4}$$
- use elementhood notation $${x | P(x)}$$
Velleman suggests that it is common to replace the $x$ in front of the vertical line with a more complex expression. He uses the example of $S$, the set of perfect squares
$$S = {n^2 | n in Bbb{N}}tag{1}$$
He claims that (1) can be written as
$$S = {x | exists n in Bbb{N}(x=n^2)}tag{2}$$
Thus
$$S = {n^2 | n in Bbb{N}} = {x | exists n in Bbb{N}(x=n^2)}$$
At this point I'm still with Velleman. Replacing x before the vertical line with a more complex expression is intuitive. Putting the statement "x is a perfect square" into the form $exists n in Bbb{N}(x=n^2)$ and defining the set S using that logical statement as an elementhood test - $S = {x | exists n in Bbb{N}(x=n^2)}$ - makes sense (this question is a good discussion. on eq 2)
Velleman continues
and therefore $x in {n^2 | n in Bbb{N}}$ means the same thing as $exists n in Bbb{N}(x = n^2)$
I am missing the jump to this final conclusion, here are my thoughts so far.
First thought
Perhaps a step Velleman is taking (and assumes reader will make) is that $x in {n^2 | n in Bbb{N}}$ is the same as $x in {x | exists n in Bbb{N}(x = n^2)}$. And then, using the following from Chapter 1.3,
in general, the statement $y in {x | P(x)}$ means the same thing as P(y)...a statement about y but not x.
from this, we can see that $x in {x | exists n in Bbb{N}(x = n^2)}$ means the same thing as $exists n in Bbb{N}(x = n^2)$, and then arrive at the conclusion.
Second thought (not entirely unrelated way of thinking about it)
$exists n in Bbb{N}(x = n^2)$ and $x in {n^2 | n in Bbb{N}}$ are both statements about x, where x is a free variable. These statement can be evaluated to true or false, but that will depend on the value of x that is used. They will always evaluate to the same truth/false value, therefore they mean the same thing.
elementary-set-theory logic proof-explanation predicate-logic
asked Jan 9 at 15:16
Jake KirschJake Kirsch
547
$endgroup$
I am reading the first two paragraphs of Chapter 2.3 in How to Prove It (Velleman) and I am unclear about why $x in {n^2 | n in Bbb{N}}$ is the same as $exists n in Bbb{N}(x = n^2)$
I will lay out Velleman's explanation and my understanding of it, I would love some feedback on if I'm thinking about this correctly.
For the reader, up until this point in the book, we know two ways to define sets.
- list elements in brackets $${1, 2, 3, 4}$$
- use elementhood notation $${x | P(x)}$$
Velleman suggests that it is common to replace the $x$ in front of the vertical line with a more complex expression. He uses the example of $S$, the set of perfect squares
$$S = {n^2 | n in Bbb{N}}tag{1}$$
He claims that (1) can be written as
$$S = {x | exists n in Bbb{N}(x=n^2)}tag{2}$$
Thus
$$S = {n^2 | n in Bbb{N}} = {x | exists n in Bbb{N}(x=n^2)}$$
At this point I'm still with Velleman. Replacing x before the vertical line with a more complex expression is intuitive. Putting the statement "x is a perfect square" into the form $exists n in Bbb{N}(x=n^2)$ and defining the set S using that logical statement as an elementhood test - $S = {x | exists n in Bbb{N}(x=n^2)}$ - makes sense (this question is a good discussion. on eq 2)
Velleman continues
and therefore $x in {n^2 | n in Bbb{N}}$ means the same thing as $exists n in Bbb{N}(x = n^2)$
I am missing the jump to this final conclusion, here are my thoughts so far.
First thought
Perhaps a step Velleman is taking (and assumes reader will make) is that $x in {n^2 | n in Bbb{N}}$ is the same as $x in {x | exists n in Bbb{N}(x = n^2)}$. And then, using the following from Chapter 1.3,
in general, the statement $y in {x | P(x)}$ means the same thing as P(y)...a statement about y but not x.
from this, we can see that $x in {x | exists n in Bbb{N}(x = n^2)}$ means the same thing as $exists n in Bbb{N}(x = n^2)$, and then arrive at the conclusion.
Second thought (not entirely unrelated way of thinking about it)
$exists n in Bbb{N}(x = n^2)$ and $x in {n^2 | n in Bbb{N}}$ are both statements about x, where x is a free variable. These statement can be evaluated to true or false, but that will depend on the value of x that is used. They will always evaluate to the same truth/false value, therefore they mean the same thing.
elementary-set-theory logic proof-explanation predicate-logic
elementary-set-theory logic proof-explanation predicate-logic
asked Jan 9 at 15:16
Jake KirschJake Kirsch
547
asked Jan 9 at 15:16
Jake KirschJake Kirsch
547
asked Jan 9 at 15:16
Jake KirschJake Kirsch
547
asked Jan 9 at 15:16
Jake KirschJake Kirsch
547
asked Jan 9 at 15:16
Jake KirschJake Kirsch
547
547
$begingroup$
Surely (2) is simply Velleman's definition of what (1) means? Then it all makes sense, right? I don't have Velleman's book to hand, so I can't check this for myself.
$endgroup$
– TonyK
Jan 9 at 15:26
$begingroup$
I would say your first and second thought are both correct, and are in fact the same thought stated in different words.
$endgroup$
– Mees de Vries
Jan 9 at 15:27
$begingroup$
@TonyK it does make sense and I agree they are equal, but I am very new to this topic and I haven't studied math in a long time! I am trying to understand exactly why the definitions are the same, based on the concepts presented in the text thus far.
$endgroup$
– Jake Kirsch
Jan 9 at 15:31
$begingroup$
You have completely misunderstood my comment! Please read it more carefully.
$endgroup$
– TonyK
Jan 9 at 15:34
$begingroup$
We have that $x in { z mid P(z) } text { iff } P(x)$.
$endgroup$
– Mauro ALLEGRANZA
Jan 9 at 15:48
|
show 2 more comments
$begingroup$
Surely (2) is simply Velleman's definition of what (1) means? Then it all makes sense, right? I don't have Velleman's book to hand, so I can't check this for myself.
$endgroup$
– TonyK
Jan 9 at 15:26
$begingroup$
I would say your first and second thought are both correct, and are in fact the same thought stated in different words.
$endgroup$
– Mees de Vries
Jan 9 at 15:27
$begingroup$
@TonyK it does make sense and I agree they are equal, but I am very new to this topic and I haven't studied math in a long time! I am trying to understand exactly why the definitions are the same, based on the concepts presented in the text thus far.
$endgroup$
– Jake Kirsch
Jan 9 at 15:31
$begingroup$
You have completely misunderstood my comment! Please read it more carefully.
$endgroup$
– TonyK
Jan 9 at 15:34
$begingroup$
We have that $x in { z mid P(z) } text { iff } P(x)$.
$endgroup$
– Mauro ALLEGRANZA
Jan 9 at 15:48
$begingroup$
Surely (2) is simply Velleman's definition of what (1) means? Then it all makes sense, right? I don't have Velleman's book to hand, so I can't check this for myself.
$endgroup$
– TonyK
Jan 9 at 15:26
$begingroup$
Surely (2) is simply Velleman's definition of what (1) means? Then it all makes sense, right? I don't have Velleman's book to hand, so I can't check this for myself.
$endgroup$
– TonyK
Jan 9 at 15:26
$begingroup$
I would say your first and second thought are both correct, and are in fact the same thought stated in different words.
$endgroup$
– Mees de Vries
Jan 9 at 15:27
$begingroup$
I would say your first and second thought are both correct, and are in fact the same thought stated in different words.
$endgroup$
– Mees de Vries
Jan 9 at 15:27
$begingroup$
@TonyK it does make sense and I agree they are equal, but I am very new to this topic and I haven't studied math in a long time! I am trying to understand exactly why the definitions are the same, based on the concepts presented in the text thus far.
$endgroup$
– Jake Kirsch
Jan 9 at 15:31
$begingroup$
@TonyK it does make sense and I agree they are equal, but I am very new to this topic and I haven't studied math in a long time! I am trying to understand exactly why the definitions are the same, based on the concepts presented in the text thus far.
$endgroup$
– Jake Kirsch
Jan 9 at 15:31
$begingroup$
You have completely misunderstood my comment! Please read it more carefully.
$endgroup$
– TonyK
Jan 9 at 15:34
$begingroup$
You have completely misunderstood my comment! Please read it more carefully.
$endgroup$
– TonyK
Jan 9 at 15:34
$begingroup$
We have that $x in { z mid P(z) } text { iff } P(x)$.
$endgroup$
– Mauro ALLEGRANZA
Jan 9 at 15:48
$begingroup$
We have that $x in { z mid P(z) } text { iff } P(x)$.
$endgroup$
– Mauro ALLEGRANZA
Jan 9 at 15:48
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
This is an intuitive explanation. Explicitly ${n^2 | n in Bbb{N}}$ is the set of all squares ${0,1,2^2,3^2,4^2,ldots}$. If $x in {n^2 | n in Bbb{N}}$ then that means $x$ is one of the squares, it can be $0$ or $1$ or $2^2$ or $3^2,4^2,ldots$ We don't know exactly which one of them but we know at least the shape that $x$ must have. $x$ must be the square of something. That's what we formalize as $exists nin mathbb{N}$ such that $x=n^2$.
Conversely if an element is of the form $x=n^2$ i.e. $exists ninmathbb{N}$ such that $x=n^2$ ($n$ can be $0,1,2,3,ldots$) then $x in {n^2 | n in Bbb{N}}$.
Thus $x in {n^2 | n in Bbb{N}}$ is the same as $exists nin mathbb{N} (x=n^2)$ (the parenthesis here reads "such that").
answered Jan 9 at 15:55
moutheticsmouthetics
50137
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3067562%2fclarification-on-set-definitions-from-how-to-prove-it-velleman-chapter-2-3%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is an intuitive explanation. Explicitly ${n^2 | n in Bbb{N}}$ is the set of all squares ${0,1,2^2,3^2,4^2,ldots}$. If $x in {n^2 | n in Bbb{N}}$ then that means $x$ is one of the squares, it can be $0$ or $1$ or $2^2$ or $3^2,4^2,ldots$ We don't know exactly which one of them but we know at least the shape that $x$ must have. $x$ must be the square of something. That's what we formalize as $exists nin mathbb{N}$ such that $x=n^2$.
Conversely if an element is of the form $x=n^2$ i.e. $exists ninmathbb{N}$ such that $x=n^2$ ($n$ can be $0,1,2,3,ldots$) then $x in {n^2 | n in Bbb{N}}$.
Thus $x in {n^2 | n in Bbb{N}}$ is the same as $exists nin mathbb{N} (x=n^2)$ (the parenthesis here reads "such that").
answered Jan 9 at 15:55
moutheticsmouthetics
50137
$endgroup$
add a comment |
$begingroup$
This is an intuitive explanation. Explicitly ${n^2 | n in Bbb{N}}$ is the set of all squares ${0,1,2^2,3^2,4^2,ldots}$. If $x in {n^2 | n in Bbb{N}}$ then that means $x$ is one of the squares, it can be $0$ or $1$ or $2^2$ or $3^2,4^2,ldots$ We don't know exactly which one of them but we know at least the shape that $x$ must have. $x$ must be the square of something. That's what we formalize as $exists nin mathbb{N}$ such that $x=n^2$.
Conversely if an element is of the form $x=n^2$ i.e. $exists ninmathbb{N}$ such that $x=n^2$ ($n$ can be $0,1,2,3,ldots$) then $x in {n^2 | n in Bbb{N}}$.
Thus $x in {n^2 | n in Bbb{N}}$ is the same as $exists nin mathbb{N} (x=n^2)$ (the parenthesis here reads "such that").
answered Jan 9 at 15:55
moutheticsmouthetics
50137
$endgroup$
add a comment |
$begingroup$
This is an intuitive explanation. Explicitly ${n^2 | n in Bbb{N}}$ is the set of all squares ${0,1,2^2,3^2,4^2,ldots}$. If $x in {n^2 | n in Bbb{N}}$ then that means $x$ is one of the squares, it can be $0$ or $1$ or $2^2$ or $3^2,4^2,ldots$ We don't know exactly which one of them but we know at least the shape that $x$ must have. $x$ must be the square of something. That's what we formalize as $exists nin mathbb{N}$ such that $x=n^2$.
Conversely if an element is of the form $x=n^2$ i.e. $exists ninmathbb{N}$ such that $x=n^2$ ($n$ can be $0,1,2,3,ldots$) then $x in {n^2 | n in Bbb{N}}$.
Thus $x in {n^2 | n in Bbb{N}}$ is the same as $exists nin mathbb{N} (x=n^2)$ (the parenthesis here reads "such that").
answered Jan 9 at 15:55
moutheticsmouthetics
50137
$endgroup$
This is an intuitive explanation. Explicitly ${n^2 | n in Bbb{N}}$ is the set of all squares ${0,1,2^2,3^2,4^2,ldots}$. If $x in {n^2 | n in Bbb{N}}$ then that means $x$ is one of the squares, it can be $0$ or $1$ or $2^2$ or $3^2,4^2,ldots$ We don't know exactly which one of them but we know at least the shape that $x$ must have. $x$ must be the square of something. That's what we formalize as $exists nin mathbb{N}$ such that $x=n^2$.
Conversely if an element is of the form $x=n^2$ i.e. $exists ninmathbb{N}$ such that $x=n^2$ ($n$ can be $0,1,2,3,ldots$) then $x in {n^2 | n in Bbb{N}}$.
Thus $x in {n^2 | n in Bbb{N}}$ is the same as $exists nin mathbb{N} (x=n^2)$ (the parenthesis here reads "such that").
answered Jan 9 at 15:55
moutheticsmouthetics
50137
answered Jan 9 at 15:55
moutheticsmouthetics
50137
answered Jan 9 at 15:55
moutheticsmouthetics
50137
answered Jan 9 at 15:55
moutheticsmouthetics
50137
50137
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3067562%2fclarification-on-set-definitions-from-how-to-prove-it-velleman-chapter-2-3%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Surely (2) is simply Velleman's definition of what (1) means? Then it all makes sense, right? I don't have Velleman's book to hand, so I can't check this for myself.
$endgroup$
– TonyK
Jan 9 at 15:26
$begingroup$
I would say your first and second thought are both correct, and are in fact the same thought stated in different words.
$endgroup$
– Mees de Vries
Jan 9 at 15:27
$begingroup$
@TonyK it does make sense and I agree they are equal, but I am very new to this topic and I haven't studied math in a long time! I am trying to understand exactly why the definitions are the same, based on the concepts presented in the text thus far.
$endgroup$
– Jake Kirsch
Jan 9 at 15:31
$begingroup$
You have completely misunderstood my comment! Please read it more carefully.
$endgroup$
– TonyK
Jan 9 at 15:34
$begingroup$
We have that $x in { z mid P(z) } text { iff } P(x)$.
$endgroup$
– Mauro ALLEGRANZA
Jan 9 at 15:48