Elementary set theory problems and “proof techniques”
$begingroup$
Let $f:Ato B$ be a function.. Consider the following functions on the powersets of $A$ and $B$:
$$f[-]:mathcal P(A)tomathcal P(B), f[S] = {yin B | exists xin S, f(x)=y}$$
$$f^{-1}[-]:mathcal P(B)tomathcal P(A), f^{-1}[T] = {xin A | f(x)in T}$$
Show that for every $S,S'subseteq A, T,T' subseteq B$:
$Ssubseteq f^{-1}[f[S]]$ and $f[f^{-1}[T]]subseteq T$
My attempt:
$f^{-1}[f[S]]={xin A|f(x)in{yin B|exists xin S, f(x)=y}}$. And this for me looks pretty obvious that $S$ must be included in this set, but how do I really prove it? What's there to say about it?
elementary-set-theory
edited Jan 17 at 19:04
Andrés E. Caicedo
65.4k8158249
asked Jan 17 at 18:52
C. CristiC. Cristi
1,629218
$endgroup$
add a comment |
$begingroup$
Let $f:Ato B$ be a function.. Consider the following functions on the powersets of $A$ and $B$:
$$f[-]:mathcal P(A)tomathcal P(B), f[S] = {yin B | exists xin S, f(x)=y}$$
$$f^{-1}[-]:mathcal P(B)tomathcal P(A), f^{-1}[T] = {xin A | f(x)in T}$$
Show that for every $S,S'subseteq A, T,T' subseteq B$:
$Ssubseteq f^{-1}[f[S]]$ and $f[f^{-1}[T]]subseteq T$
My attempt:
$f^{-1}[f[S]]={xin A|f(x)in{yin B|exists xin S, f(x)=y}}$. And this for me looks pretty obvious that $S$ must be included in this set, but how do I really prove it? What's there to say about it?
elementary-set-theory
edited Jan 17 at 19:04
Andrés E. Caicedo
65.4k8158249
asked Jan 17 at 18:52
C. CristiC. Cristi
1,629218
$endgroup$
$begingroup$
The standard way to prove that $Ssubset Z$ (for any set $Z$) is to prove, for all $x$, that $xin S implies xin Z$. So you need to prove the following statement: if $xin S$, then $xin f^{-1}[f[S]]$. Can you do that from the definition you wrote down? (or rather, the corrected definition where you don't use $x$ to mean two different things)
$endgroup$
– Greg Martin
Jan 17 at 19:00
$begingroup$
@GregMartin I don't think, I mean, not really, I don't think I can think this through, so, $f[S]$ is the set with all $yin B$ such that $f(x)=y$ and $x$ is $in S$, and $f^{-1}[f[S]]$ is the set with all $x in A$ such that $f(x) in f[S]$ so all $x$ in $A$ not $S$ and I don't get it
$endgroup$
– C. Cristi
Jan 17 at 20:04
add a comment |
$begingroup$
Let $f:Ato B$ be a function.. Consider the following functions on the powersets of $A$ and $B$:
$$f[-]:mathcal P(A)tomathcal P(B), f[S] = {yin B | exists xin S, f(x)=y}$$
$$f^{-1}[-]:mathcal P(B)tomathcal P(A), f^{-1}[T] = {xin A | f(x)in T}$$
Show that for every $S,S'subseteq A, T,T' subseteq B$:
$Ssubseteq f^{-1}[f[S]]$ and $f[f^{-1}[T]]subseteq T$
My attempt:
$f^{-1}[f[S]]={xin A|f(x)in{yin B|exists xin S, f(x)=y}}$. And this for me looks pretty obvious that $S$ must be included in this set, but how do I really prove it? What's there to say about it?
elementary-set-theory
edited Jan 17 at 19:04
Andrés E. Caicedo
65.4k8158249
asked Jan 17 at 18:52
C. CristiC. Cristi
1,629218
$endgroup$
Let $f:Ato B$ be a function.. Consider the following functions on the powersets of $A$ and $B$:
$$f[-]:mathcal P(A)tomathcal P(B), f[S] = {yin B | exists xin S, f(x)=y}$$
$$f^{-1}[-]:mathcal P(B)tomathcal P(A), f^{-1}[T] = {xin A | f(x)in T}$$
Show that for every $S,S'subseteq A, T,T' subseteq B$:
$Ssubseteq f^{-1}[f[S]]$ and $f[f^{-1}[T]]subseteq T$
My attempt:
$f^{-1}[f[S]]={xin A|f(x)in{yin B|exists xin S, f(x)=y}}$. And this for me looks pretty obvious that $S$ must be included in this set, but how do I really prove it? What's there to say about it?
elementary-set-theory
elementary-set-theory
edited Jan 17 at 19:04
Andrés E. Caicedo
65.4k8158249
asked Jan 17 at 18:52
C. CristiC. Cristi
1,629218
edited Jan 17 at 19:04
Andrés E. Caicedo
65.4k8158249
asked Jan 17 at 18:52
C. CristiC. Cristi
1,629218
edited Jan 17 at 19:04
Andrés E. Caicedo
65.4k8158249
edited Jan 17 at 19:04
Andrés E. Caicedo
65.4k8158249
edited Jan 17 at 19:04
Andrés E. Caicedo
65.4k8158249
65.4k8158249
asked Jan 17 at 18:52
C. CristiC. Cristi
1,629218
asked Jan 17 at 18:52
C. CristiC. Cristi
1,629218
asked Jan 17 at 18:52
C. CristiC. Cristi
1,629218
1,629218
$begingroup$
The standard way to prove that $Ssubset Z$ (for any set $Z$) is to prove, for all $x$, that $xin S implies xin Z$. So you need to prove the following statement: if $xin S$, then $xin f^{-1}[f[S]]$. Can you do that from the definition you wrote down? (or rather, the corrected definition where you don't use $x$ to mean two different things)
$endgroup$
– Greg Martin
Jan 17 at 19:00
$begingroup$
@GregMartin I don't think, I mean, not really, I don't think I can think this through, so, $f[S]$ is the set with all $yin B$ such that $f(x)=y$ and $x$ is $in S$, and $f^{-1}[f[S]]$ is the set with all $x in A$ such that $f(x) in f[S]$ so all $x$ in $A$ not $S$ and I don't get it
$endgroup$
– C. Cristi
Jan 17 at 20:04
add a comment |
$begingroup$
The standard way to prove that $Ssubset Z$ (for any set $Z$) is to prove, for all $x$, that $xin S implies xin Z$. So you need to prove the following statement: if $xin S$, then $xin f^{-1}[f[S]]$. Can you do that from the definition you wrote down? (or rather, the corrected definition where you don't use $x$ to mean two different things)
$endgroup$
– Greg Martin
Jan 17 at 19:00
$begingroup$
@GregMartin I don't think, I mean, not really, I don't think I can think this through, so, $f[S]$ is the set with all $yin B$ such that $f(x)=y$ and $x$ is $in S$, and $f^{-1}[f[S]]$ is the set with all $x in A$ such that $f(x) in f[S]$ so all $x$ in $A$ not $S$ and I don't get it
$endgroup$
– C. Cristi
Jan 17 at 20:04
$begingroup$
The standard way to prove that $Ssubset Z$ (for any set $Z$) is to prove, for all $x$, that $xin S implies xin Z$. So you need to prove the following statement: if $xin S$, then $xin f^{-1}[f[S]]$. Can you do that from the definition you wrote down? (or rather, the corrected definition where you don't use $x$ to mean two different things)
$endgroup$
– Greg Martin
Jan 17 at 19:00
$begingroup$
The standard way to prove that $Ssubset Z$ (for any set $Z$) is to prove, for all $x$, that $xin S implies xin Z$. So you need to prove the following statement: if $xin S$, then $xin f^{-1}[f[S]]$. Can you do that from the definition you wrote down? (or rather, the corrected definition where you don't use $x$ to mean two different things)
$endgroup$
– Greg Martin
Jan 17 at 19:00
$begingroup$
@GregMartin I don't think, I mean, not really, I don't think I can think this through, so, $f[S]$ is the set with all $yin B$ such that $f(x)=y$ and $x$ is $in S$, and $f^{-1}[f[S]]$ is the set with all $x in A$ such that $f(x) in f[S]$ so all $x$ in $A$ not $S$ and I don't get it
$endgroup$
– C. Cristi
Jan 17 at 20:04
$begingroup$
@GregMartin I don't think, I mean, not really, I don't think I can think this through, so, $f[S]$ is the set with all $yin B$ such that $f(x)=y$ and $x$ is $in S$, and $f^{-1}[f[S]]$ is the set with all $x in A$ such that $f(x) in f[S]$ so all $x$ in $A$ not $S$ and I don't get it
$endgroup$
– C. Cristi
Jan 17 at 20:04
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I am a bit confused about your notation.
Is this version correct?
Assume $f:A to B$. Define $g:P(A) to P(B)$, $S to f^{-1}(S)$.
$h:P(B) to P(A)$, $S to f(S)$. Show $S subseteq hg(S)$, $gh(S) subseteq S$.
$S subset$ $f^{-1}(f(S)) = hg(S)$. Proof.
If $x in Simplies f(x) in f(S)$, $x in f^{-1}(f(S))$.
$f(f^{-1}(S)) = gh(S)$ $subset S$. Proof.
If $y in f(f^{-1}(S))implies exists x in f^{-1}(S)$ with $y = f(x)$;
$exists x$ with $f(x) in S$ and $y = f(x)implies y in S$.
edited Jan 18 at 12:09
Thomas Shelby
3,1271524
answered Jan 18 at 0:38
William ElliotWilliam Elliot
8,1212720
$endgroup$
$begingroup$
looks right I think, but can you edit more using mathJax?
$endgroup$
– C. Cristi
Jan 18 at 11:55
$begingroup$
Review my edit please.
$endgroup$
– C. Cristi
Jan 18 at 11:59
$begingroup$
Can you add more details? You just apply $f^{-1}$ on both sides at the first proof?
$endgroup$
– C. Cristi
Jan 18 at 12:02
$begingroup$
No I did not do that to both sides, is used the definitions of f and inverse f as extended for sets. @C.Cristi
$endgroup$
– William Elliot
Jan 18 at 12:28
$begingroup$
Remove the implication symbols and replace with the orginal English as they mess up the logic. If ... implies ... is absurd. @C.Cristi
$endgroup$
– William Elliot
Jan 18 at 12:29
add a comment |
$begingroup$
Here is how I would do these proofs: using element chasing, so seeing which elements are part of the most complex side, and then expanding definitions and simplifying towards the simpler side.$%
require{begingroup}
begingroup
newcommand{calc}{begin{align} quad &}
newcommand{op}[1]{\ #1 quad & quad unicode{x201c}}
newcommand{hints}[1]{mbox{#1} \ quad & quad phantom{unicode{x201c}} }
newcommand{hint}[1]{mbox{#1} unicode{x201d} \ quad & }
newcommand{endcalc}{end{align}}
newcommand{subcalch}[1]{\ quad & quad #1 \ quad &}
newcommand{subcalc}{quad begin{aligned} quad & \ bullet quad & }
newcommand{endsubcalc}{end{aligned} \ \ cdot quad &}
newcommand{Ref}[1]{text{(#1)}}
newcommand{then}{implies}
newcommand{when}{impliedby}
newcommand{equiv}{iff}
newcommand{true}{text{true}}
%$
For the first problem, let's see which $;s;$ are in the right hand side:
$$calc
s in f^{-1}[f[S]]
opequivhint{definition of $;f^{-1}[-];$}
s in A ;land; f(s) in f[S]
opequivhint{definition of $;f[-];$}
s in A ;land; f(s) in B ;land; exists x in S, f(x) = f(s)
opwhenhint{left: use $;S subseteq A;$; middle: $;f : A to B;$; right: choose $;x:=s;$}
s in S ;land; true ;land; f(s) = f(s)
opequivhint{logic: simplify}
s in S
endcalc$$
which (by set extensionality) proves $;S subseteq f^{-1}[f[S]];$.
The second statement can be proven similarly.
$%
endgroup
%$
answered Jan 19 at 10:17
Marnix KloosterMarnix Klooster
4,21322147
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I am a bit confused about your notation.
Is this version correct?
Assume $f:A to B$. Define $g:P(A) to P(B)$, $S to f^{-1}(S)$.
$h:P(B) to P(A)$, $S to f(S)$. Show $S subseteq hg(S)$, $gh(S) subseteq S$.
$S subset$ $f^{-1}(f(S)) = hg(S)$. Proof.
If $x in Simplies f(x) in f(S)$, $x in f^{-1}(f(S))$.
$f(f^{-1}(S)) = gh(S)$ $subset S$. Proof.
If $y in f(f^{-1}(S))implies exists x in f^{-1}(S)$ with $y = f(x)$;
$exists x$ with $f(x) in S$ and $y = f(x)implies y in S$.
edited Jan 18 at 12:09
Thomas Shelby
3,1271524
answered Jan 18 at 0:38
William ElliotWilliam Elliot
8,1212720
$endgroup$
$begingroup$
looks right I think, but can you edit more using mathJax?
$endgroup$
– C. Cristi
Jan 18 at 11:55
$begingroup$
Review my edit please.
$endgroup$
– C. Cristi
Jan 18 at 11:59
$begingroup$
Can you add more details? You just apply $f^{-1}$ on both sides at the first proof?
$endgroup$
– C. Cristi
Jan 18 at 12:02
$begingroup$
No I did not do that to both sides, is used the definitions of f and inverse f as extended for sets. @C.Cristi
$endgroup$
– William Elliot
Jan 18 at 12:28
$begingroup$
Remove the implication symbols and replace with the orginal English as they mess up the logic. If ... implies ... is absurd. @C.Cristi
$endgroup$
– William Elliot
Jan 18 at 12:29
add a comment |
$begingroup$
I am a bit confused about your notation.
Is this version correct?
Assume $f:A to B$. Define $g:P(A) to P(B)$, $S to f^{-1}(S)$.
$h:P(B) to P(A)$, $S to f(S)$. Show $S subseteq hg(S)$, $gh(S) subseteq S$.
$S subset$ $f^{-1}(f(S)) = hg(S)$. Proof.
If $x in Simplies f(x) in f(S)$, $x in f^{-1}(f(S))$.
$f(f^{-1}(S)) = gh(S)$ $subset S$. Proof.
If $y in f(f^{-1}(S))implies exists x in f^{-1}(S)$ with $y = f(x)$;
$exists x$ with $f(x) in S$ and $y = f(x)implies y in S$.
edited Jan 18 at 12:09
Thomas Shelby
3,1271524
answered Jan 18 at 0:38
William ElliotWilliam Elliot
8,1212720
$endgroup$
$begingroup$
looks right I think, but can you edit more using mathJax?
$endgroup$
– C. Cristi
Jan 18 at 11:55
$begingroup$
Review my edit please.
$endgroup$
– C. Cristi
Jan 18 at 11:59
$begingroup$
Can you add more details? You just apply $f^{-1}$ on both sides at the first proof?
$endgroup$
– C. Cristi
Jan 18 at 12:02
$begingroup$
No I did not do that to both sides, is used the definitions of f and inverse f as extended for sets. @C.Cristi
$endgroup$
– William Elliot
Jan 18 at 12:28
$begingroup$
Remove the implication symbols and replace with the orginal English as they mess up the logic. If ... implies ... is absurd. @C.Cristi
$endgroup$
– William Elliot
Jan 18 at 12:29
add a comment |
$begingroup$
I am a bit confused about your notation.
Is this version correct?
Assume $f:A to B$. Define $g:P(A) to P(B)$, $S to f^{-1}(S)$.
$h:P(B) to P(A)$, $S to f(S)$. Show $S subseteq hg(S)$, $gh(S) subseteq S$.
$S subset$ $f^{-1}(f(S)) = hg(S)$. Proof.
If $x in Simplies f(x) in f(S)$, $x in f^{-1}(f(S))$.
$f(f^{-1}(S)) = gh(S)$ $subset S$. Proof.
If $y in f(f^{-1}(S))implies exists x in f^{-1}(S)$ with $y = f(x)$;
$exists x$ with $f(x) in S$ and $y = f(x)implies y in S$.
edited Jan 18 at 12:09
Thomas Shelby
3,1271524
answered Jan 18 at 0:38
William ElliotWilliam Elliot
8,1212720
$endgroup$
I am a bit confused about your notation.
Is this version correct?
Assume $f:A to B$. Define $g:P(A) to P(B)$, $S to f^{-1}(S)$.
$h:P(B) to P(A)$, $S to f(S)$. Show $S subseteq hg(S)$, $gh(S) subseteq S$.
$S subset$ $f^{-1}(f(S)) = hg(S)$. Proof.
If $x in Simplies f(x) in f(S)$, $x in f^{-1}(f(S))$.
$f(f^{-1}(S)) = gh(S)$ $subset S$. Proof.
If $y in f(f^{-1}(S))implies exists x in f^{-1}(S)$ with $y = f(x)$;
$exists x$ with $f(x) in S$ and $y = f(x)implies y in S$.
edited Jan 18 at 12:09
Thomas Shelby
3,1271524
answered Jan 18 at 0:38
William ElliotWilliam Elliot
8,1212720
edited Jan 18 at 12:09
Thomas Shelby
3,1271524
edited Jan 18 at 12:09
Thomas Shelby
3,1271524
edited Jan 18 at 12:09
Thomas Shelby
3,1271524
3,1271524
answered Jan 18 at 0:38
William ElliotWilliam Elliot
8,1212720
answered Jan 18 at 0:38
William ElliotWilliam Elliot
8,1212720
answered Jan 18 at 0:38
William ElliotWilliam Elliot
8,1212720
8,1212720
$begingroup$
looks right I think, but can you edit more using mathJax?
$endgroup$
– C. Cristi
Jan 18 at 11:55
$begingroup$
Review my edit please.
$endgroup$
– C. Cristi
Jan 18 at 11:59
$begingroup$
Can you add more details? You just apply $f^{-1}$ on both sides at the first proof?
$endgroup$
– C. Cristi
Jan 18 at 12:02
$begingroup$
No I did not do that to both sides, is used the definitions of f and inverse f as extended for sets. @C.Cristi
$endgroup$
– William Elliot
Jan 18 at 12:28
$begingroup$
Remove the implication symbols and replace with the orginal English as they mess up the logic. If ... implies ... is absurd. @C.Cristi
$endgroup$
– William Elliot
Jan 18 at 12:29
add a comment |
$begingroup$
looks right I think, but can you edit more using mathJax?
$endgroup$
– C. Cristi
Jan 18 at 11:55
$begingroup$
Review my edit please.
$endgroup$
– C. Cristi
Jan 18 at 11:59
$begingroup$
Can you add more details? You just apply $f^{-1}$ on both sides at the first proof?
$endgroup$
– C. Cristi
Jan 18 at 12:02
$begingroup$
No I did not do that to both sides, is used the definitions of f and inverse f as extended for sets. @C.Cristi
$endgroup$
– William Elliot
Jan 18 at 12:28
$begingroup$
Remove the implication symbols and replace with the orginal English as they mess up the logic. If ... implies ... is absurd. @C.Cristi
$endgroup$
– William Elliot
Jan 18 at 12:29
$begingroup$
looks right I think, but can you edit more using mathJax?
$endgroup$
– C. Cristi
Jan 18 at 11:55
$begingroup$
looks right I think, but can you edit more using mathJax?
$endgroup$
– C. Cristi
Jan 18 at 11:55
$begingroup$
Review my edit please.
$endgroup$
– C. Cristi
Jan 18 at 11:59
$begingroup$
Review my edit please.
$endgroup$
– C. Cristi
Jan 18 at 11:59
$begingroup$
Can you add more details? You just apply $f^{-1}$ on both sides at the first proof?
$endgroup$
– C. Cristi
Jan 18 at 12:02
$begingroup$
Can you add more details? You just apply $f^{-1}$ on both sides at the first proof?
$endgroup$
– C. Cristi
Jan 18 at 12:02
$begingroup$
No I did not do that to both sides, is used the definitions of f and inverse f as extended for sets. @C.Cristi
$endgroup$
– William Elliot
Jan 18 at 12:28
$begingroup$
No I did not do that to both sides, is used the definitions of f and inverse f as extended for sets. @C.Cristi
$endgroup$
– William Elliot
Jan 18 at 12:28
$begingroup$
Remove the implication symbols and replace with the orginal English as they mess up the logic. If ... implies ... is absurd. @C.Cristi
$endgroup$
– William Elliot
Jan 18 at 12:29
$begingroup$
Remove the implication symbols and replace with the orginal English as they mess up the logic. If ... implies ... is absurd. @C.Cristi
$endgroup$
– William Elliot
Jan 18 at 12:29
add a comment |
$begingroup$
Here is how I would do these proofs: using element chasing, so seeing which elements are part of the most complex side, and then expanding definitions and simplifying towards the simpler side.$%
require{begingroup}
begingroup
newcommand{calc}{begin{align} quad &}
newcommand{op}[1]{\ #1 quad & quad unicode{x201c}}
newcommand{hints}[1]{mbox{#1} \ quad & quad phantom{unicode{x201c}} }
newcommand{hint}[1]{mbox{#1} unicode{x201d} \ quad & }
newcommand{endcalc}{end{align}}
newcommand{subcalch}[1]{\ quad & quad #1 \ quad &}
newcommand{subcalc}{quad begin{aligned} quad & \ bullet quad & }
newcommand{endsubcalc}{end{aligned} \ \ cdot quad &}
newcommand{Ref}[1]{text{(#1)}}
newcommand{then}{implies}
newcommand{when}{impliedby}
newcommand{equiv}{iff}
newcommand{true}{text{true}}
%$
For the first problem, let's see which $;s;$ are in the right hand side:
$$calc
s in f^{-1}[f[S]]
opequivhint{definition of $;f^{-1}[-];$}
s in A ;land; f(s) in f[S]
opequivhint{definition of $;f[-];$}
s in A ;land; f(s) in B ;land; exists x in S, f(x) = f(s)
opwhenhint{left: use $;S subseteq A;$; middle: $;f : A to B;$; right: choose $;x:=s;$}
s in S ;land; true ;land; f(s) = f(s)
opequivhint{logic: simplify}
s in S
endcalc$$
which (by set extensionality) proves $;S subseteq f^{-1}[f[S]];$.
The second statement can be proven similarly.
$%
endgroup
%$
answered Jan 19 at 10:17
Marnix KloosterMarnix Klooster
4,21322147
$endgroup$
add a comment |
$begingroup$
Here is how I would do these proofs: using element chasing, so seeing which elements are part of the most complex side, and then expanding definitions and simplifying towards the simpler side.$%
require{begingroup}
begingroup
newcommand{calc}{begin{align} quad &}
newcommand{op}[1]{\ #1 quad & quad unicode{x201c}}
newcommand{hints}[1]{mbox{#1} \ quad & quad phantom{unicode{x201c}} }
newcommand{hint}[1]{mbox{#1} unicode{x201d} \ quad & }
newcommand{endcalc}{end{align}}
newcommand{subcalch}[1]{\ quad & quad #1 \ quad &}
newcommand{subcalc}{quad begin{aligned} quad & \ bullet quad & }
newcommand{endsubcalc}{end{aligned} \ \ cdot quad &}
newcommand{Ref}[1]{text{(#1)}}
newcommand{then}{implies}
newcommand{when}{impliedby}
newcommand{equiv}{iff}
newcommand{true}{text{true}}
%$
For the first problem, let's see which $;s;$ are in the right hand side:
$$calc
s in f^{-1}[f[S]]
opequivhint{definition of $;f^{-1}[-];$}
s in A ;land; f(s) in f[S]
opequivhint{definition of $;f[-];$}
s in A ;land; f(s) in B ;land; exists x in S, f(x) = f(s)
opwhenhint{left: use $;S subseteq A;$; middle: $;f : A to B;$; right: choose $;x:=s;$}
s in S ;land; true ;land; f(s) = f(s)
opequivhint{logic: simplify}
s in S
endcalc$$
which (by set extensionality) proves $;S subseteq f^{-1}[f[S]];$.
The second statement can be proven similarly.
$%
endgroup
%$
answered Jan 19 at 10:17
Marnix KloosterMarnix Klooster
4,21322147
$endgroup$
add a comment |
$begingroup$
Here is how I would do these proofs: using element chasing, so seeing which elements are part of the most complex side, and then expanding definitions and simplifying towards the simpler side.$%
require{begingroup}
begingroup
newcommand{calc}{begin{align} quad &}
newcommand{op}[1]{\ #1 quad & quad unicode{x201c}}
newcommand{hints}[1]{mbox{#1} \ quad & quad phantom{unicode{x201c}} }
newcommand{hint}[1]{mbox{#1} unicode{x201d} \ quad & }
newcommand{endcalc}{end{align}}
newcommand{subcalch}[1]{\ quad & quad #1 \ quad &}
newcommand{subcalc}{quad begin{aligned} quad & \ bullet quad & }
newcommand{endsubcalc}{end{aligned} \ \ cdot quad &}
newcommand{Ref}[1]{text{(#1)}}
newcommand{then}{implies}
newcommand{when}{impliedby}
newcommand{equiv}{iff}
newcommand{true}{text{true}}
%$
For the first problem, let's see which $;s;$ are in the right hand side:
$$calc
s in f^{-1}[f[S]]
opequivhint{definition of $;f^{-1}[-];$}
s in A ;land; f(s) in f[S]
opequivhint{definition of $;f[-];$}
s in A ;land; f(s) in B ;land; exists x in S, f(x) = f(s)
opwhenhint{left: use $;S subseteq A;$; middle: $;f : A to B;$; right: choose $;x:=s;$}
s in S ;land; true ;land; f(s) = f(s)
opequivhint{logic: simplify}
s in S
endcalc$$
which (by set extensionality) proves $;S subseteq f^{-1}[f[S]];$.
The second statement can be proven similarly.
$%
endgroup
%$
answered Jan 19 at 10:17
Marnix KloosterMarnix Klooster
4,21322147
$endgroup$
Here is how I would do these proofs: using element chasing, so seeing which elements are part of the most complex side, and then expanding definitions and simplifying towards the simpler side.$%
require{begingroup}
begingroup
newcommand{calc}{begin{align} quad &}
newcommand{op}[1]{\ #1 quad & quad unicode{x201c}}
newcommand{hints}[1]{mbox{#1} \ quad & quad phantom{unicode{x201c}} }
newcommand{hint}[1]{mbox{#1} unicode{x201d} \ quad & }
newcommand{endcalc}{end{align}}
newcommand{subcalch}[1]{\ quad & quad #1 \ quad &}
newcommand{subcalc}{quad begin{aligned} quad & \ bullet quad & }
newcommand{endsubcalc}{end{aligned} \ \ cdot quad &}
newcommand{Ref}[1]{text{(#1)}}
newcommand{then}{implies}
newcommand{when}{impliedby}
newcommand{equiv}{iff}
newcommand{true}{text{true}}
%$
For the first problem, let's see which $;s;$ are in the right hand side:
$$calc
s in f^{-1}[f[S]]
opequivhint{definition of $;f^{-1}[-];$}
s in A ;land; f(s) in f[S]
opequivhint{definition of $;f[-];$}
s in A ;land; f(s) in B ;land; exists x in S, f(x) = f(s)
opwhenhint{left: use $;S subseteq A;$; middle: $;f : A to B;$; right: choose $;x:=s;$}
s in S ;land; true ;land; f(s) = f(s)
opequivhint{logic: simplify}
s in S
endcalc$$
which (by set extensionality) proves $;S subseteq f^{-1}[f[S]];$.
The second statement can be proven similarly.
$%
endgroup
%$
answered Jan 19 at 10:17
Marnix KloosterMarnix Klooster
4,21322147
answered Jan 19 at 10:17
Marnix KloosterMarnix Klooster
4,21322147
answered Jan 19 at 10:17
Marnix KloosterMarnix Klooster
4,21322147
answered Jan 19 at 10:17
Marnix KloosterMarnix Klooster
4,21322147
4,21322147
add a comment |
add a comment |
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$begingroup$
The standard way to prove that $Ssubset Z$ (for any set $Z$) is to prove, for all $x$, that $xin S implies xin Z$. So you need to prove the following statement: if $xin S$, then $xin f^{-1}[f[S]]$. Can you do that from the definition you wrote down? (or rather, the corrected definition where you don't use $x$ to mean two different things)
$endgroup$
– Greg Martin
Jan 17 at 19:00
$begingroup$
@GregMartin I don't think, I mean, not really, I don't think I can think this through, so, $f[S]$ is the set with all $yin B$ such that $f(x)=y$ and $x$ is $in S$, and $f^{-1}[f[S]]$ is the set with all $x in A$ such that $f(x) in f[S]$ so all $x$ in $A$ not $S$ and I don't get it
$endgroup$
– C. Cristi
Jan 17 at 20:04