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Prove that if |A| ≤ |B| and |B| ≤ |C| then |A| ≤ |C| (1 Viewer)

sarah666

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Hello,
Can someone help me on this question please ?​


Prove that if |A| ≤ |B| and |B| ≤ |C| then |A| ≤ |C|

Thanks in advance.
 

Trebla

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|A|≤ |B|
|B| ≤ |C|
So, |A| ≤ |B| ≤ |C|
hence |A| ≤ |C|
 

conics2008

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|A| ≤ |B| and |B| ≤ |C| then |A| ≤ |C|

You said if A is less than or equal to B and if B is less than or equal to C then

A must be less than or equal to C because if you say B = C

then A less than or equal to C ..

What kind of maths questoin is this... ?
 

lolokay

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(leaving out |'s)

A ≤ B, B≤ C
let B = A + x
let C = B + y
C = A + x + y
A≤ A+x
0≤ x
B≤ B + y
0≤ y
0≤ x≤ x + y
0≤ x + y *
A≤ A + x + y
A≤ C

*although, given this step you may as well just use A≤ B≤ C, so A≤ C

edit: deleted below post since it doesn't apply, assuming cardinality is meant
 
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Affinity

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|A| ≤ |B| could mean "there exist an injection from set A into set B"
 

sarah666

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Affinity said:
|A| ≤ |B| could mean "there exist an injection from set A into set B"
If this is the case, would the proof be any different from the proofs given by the posts above?

Also, thanks everyone for the help so far.
 

Affinity

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would be slightly different.. by the way, where did you get that question from
 

Slidey

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Look up the property of transitivity, sarah.
 

sarah666

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Affinity said:
would be slightly different.. by the way, where did you get that question from
Oh so it is different, can you please show me the proof ? Because I have absolutely no idea.

This question was given to me by a friend of mine and she doesn't know how it is done as well. :-(. If you can help me, I wolud be very appreciated.
 

darkliight

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Let f:A->B and g:B->C be injective (these exist by assumption). Then gof (g after f) is injective. To see this, suppose gof(x) = gof(y), then f(x) = f(y) since g is injective, and x = y since f is injective.

This surely isn't what your teacher wanted though, ie, your teacher probably wanted something like the first couple of posts here.
 

kooltrainer

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wahh?.. wtf is injection from B into A.. wah??
dw, not in syllabus xD
 

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