Assumed existence of an intermediate set given $A subseteq B$
0
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If I know that there are two sets, $A$ and $B$, such that $B subseteq A$, can I assume for a proof that there is a $C$ such that $B subseteq C subseteq A$ since either $A = B = C$, $B = C subset A$, $B subset C = A$, or that there exists a $C$ such that $B subset C subset A$?
elementary-set-theory
asked Jan 16 at 12:30
PixelArtDragonPixelArtDragon
1183
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add a comment |
0
$begingroup$
If I know that there are two sets, $A$ and $B$, such that $B subseteq A$, can I assume for a proof that there is a $C$ such that $B subseteq C subseteq A$ since either $A = B = C$, $B = C subset A$, $B subset C = A$, or that there exists a $C$ such that $B subset C subset A$?
elementary-set-theory
asked Jan 16 at 12:30
PixelArtDragonPixelArtDragon
1183
$endgroup$
2
$begingroup$
If $subseteq$, obviously yes : $C=A$ or $C=B$ will do.
$endgroup$
– Mauro ALLEGRANZA
Jan 16 at 12:37
$begingroup$
$C=B$ or $C=A$ always works.
$endgroup$
– xbh
Jan 16 at 12:38
$begingroup$
If $subset$, not necessarily. A set ${ a }$ with a single element (singleton) has only two subsets : $emptyset$ and ${ a }$ itself.
$endgroup$
– Mauro ALLEGRANZA
Jan 16 at 12:39
$begingroup$
@MauroALLEGRANZA The thing is, while I know that $A subseteq B$, I need to apply a function to each that doesn't necessarily preserve the equality. It turned out that if I take $C = A$ it works, but I wasn't sure before and needed the possibility of an intermediate set.
$endgroup$
– PixelArtDragon
Jan 16 at 13:06
add a comment |
0
0
0
$begingroup$
If I know that there are two sets, $A$ and $B$, such that $B subseteq A$, can I assume for a proof that there is a $C$ such that $B subseteq C subseteq A$ since either $A = B = C$, $B = C subset A$, $B subset C = A$, or that there exists a $C$ such that $B subset C subset A$?
elementary-set-theory
asked Jan 16 at 12:30
PixelArtDragonPixelArtDragon
1183
$endgroup$
If I know that there are two sets, $A$ and $B$, such that $B subseteq A$, can I assume for a proof that there is a $C$ such that $B subseteq C subseteq A$ since either $A = B = C$, $B = C subset A$, $B subset C = A$, or that there exists a $C$ such that $B subset C subset A$?
elementary-set-theory
elementary-set-theory
asked Jan 16 at 12:30
PixelArtDragonPixelArtDragon
1183
asked Jan 16 at 12:30
PixelArtDragonPixelArtDragon
1183
asked Jan 16 at 12:30
PixelArtDragonPixelArtDragon
1183
asked Jan 16 at 12:30
PixelArtDragonPixelArtDragon
1183
asked Jan 16 at 12:30
PixelArtDragonPixelArtDragon
1183
1183
2
$begingroup$
If $subseteq$, obviously yes : $C=A$ or $C=B$ will do.
$endgroup$
– Mauro ALLEGRANZA
Jan 16 at 12:37
$begingroup$
$C=B$ or $C=A$ always works.
$endgroup$
– xbh
Jan 16 at 12:38
$begingroup$
If $subset$, not necessarily. A set ${ a }$ with a single element (singleton) has only two subsets : $emptyset$ and ${ a }$ itself.
$endgroup$
– Mauro ALLEGRANZA
Jan 16 at 12:39
$begingroup$
@MauroALLEGRANZA The thing is, while I know that $A subseteq B$, I need to apply a function to each that doesn't necessarily preserve the equality. It turned out that if I take $C = A$ it works, but I wasn't sure before and needed the possibility of an intermediate set.
$endgroup$
– PixelArtDragon
Jan 16 at 13:06
add a comment |
2
$begingroup$
If $subseteq$, obviously yes : $C=A$ or $C=B$ will do.
$endgroup$
– Mauro ALLEGRANZA
Jan 16 at 12:37
$begingroup$
$C=B$ or $C=A$ always works.
$endgroup$
– xbh
Jan 16 at 12:38
$begingroup$
If $subset$, not necessarily. A set ${ a }$ with a single element (singleton) has only two subsets : $emptyset$ and ${ a }$ itself.
$endgroup$
– Mauro ALLEGRANZA
Jan 16 at 12:39
$begingroup$
@MauroALLEGRANZA The thing is, while I know that $A subseteq B$, I need to apply a function to each that doesn't necessarily preserve the equality. It turned out that if I take $C = A$ it works, but I wasn't sure before and needed the possibility of an intermediate set.
$endgroup$
– PixelArtDragon
Jan 16 at 13:06
2
2
$begingroup$
If $subseteq$, obviously yes : $C=A$ or $C=B$ will do.
$endgroup$
– Mauro ALLEGRANZA
Jan 16 at 12:37
$begingroup$
If $subseteq$, obviously yes : $C=A$ or $C=B$ will do.
$endgroup$
– Mauro ALLEGRANZA
Jan 16 at 12:37
$begingroup$
$C=B$ or $C=A$ always works.
$endgroup$
– xbh
Jan 16 at 12:38
$begingroup$
$C=B$ or $C=A$ always works.
$endgroup$
– xbh
Jan 16 at 12:38
$begingroup$
If $subset$, not necessarily. A set ${ a }$ with a single element (singleton) has only two subsets : $emptyset$ and ${ a }$ itself.
$endgroup$
– Mauro ALLEGRANZA
Jan 16 at 12:39
$begingroup$
If $subset$, not necessarily. A set ${ a }$ with a single element (singleton) has only two subsets : $emptyset$ and ${ a }$ itself.
$endgroup$
– Mauro ALLEGRANZA
Jan 16 at 12:39
$begingroup$
@MauroALLEGRANZA The thing is, while I know that $A subseteq B$, I need to apply a function to each that doesn't necessarily preserve the equality. It turned out that if I take $C = A$ it works, but I wasn't sure before and needed the possibility of an intermediate set.
$endgroup$
– PixelArtDragon
Jan 16 at 13:06
$begingroup$
@MauroALLEGRANZA The thing is, while I know that $A subseteq B$, I need to apply a function to each that doesn't necessarily preserve the equality. It turned out that if I take $C = A$ it works, but I wasn't sure before and needed the possibility of an intermediate set.
$endgroup$
– PixelArtDragon
Jan 16 at 13:06
add a comment |
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2
$begingroup$
If $subseteq$, obviously yes : $C=A$ or $C=B$ will do.
$endgroup$
– Mauro ALLEGRANZA
Jan 16 at 12:37
$begingroup$
$C=B$ or $C=A$ always works.
$endgroup$
– xbh
Jan 16 at 12:38
$begingroup$
If $subset$, not necessarily. A set ${ a }$ with a single element (singleton) has only two subsets : $emptyset$ and ${ a }$ itself.
$endgroup$
– Mauro ALLEGRANZA
Jan 16 at 12:39
$begingroup$
@MauroALLEGRANZA The thing is, while I know that $A subseteq B$, I need to apply a function to each that doesn't necessarily preserve the equality. It turned out that if I take $C = A$ it works, but I wasn't sure before and needed the possibility of an intermediate set.
$endgroup$
– PixelArtDragon
Jan 16 at 13:06