MHB JPARK 's question at Yahoo Answers (Cardinality)

  • Thread starter Thread starter Fernando Revilla
  • Start date Start date
  • Tags Tags
    Cardinality
AI Thread Summary
To demonstrate that if |A| ≤ |B| and |B| ≤ |C|, then |A| ≤ |C|, one can use the concept of injective functions. If there exists an injective function f from set A to set B and an injective function g from set B to set C, then the composition of these functions, g ∘ f, creates an injective function from A to C. This shows that |A| is less than or equal to |C|, confirming the transitive property of cardinality. The response effectively clarifies the relationship between the sets and the implications of injective mappings. Understanding these concepts is crucial for grasping cardinality in set theory.
Fernando Revilla
Gold Member
MHB
Messages
631
Reaction score
0
Here is the question:

My professor wasn't clear in lecture today, so I'm not exactly sure how I should show this...Can anyone help?

Let A,B,C be sets. Show that if |A| ≤ |B| and |B| ≤ |C|, then |A| ≤ |C|

Here is a link to the question:

Cardinality of Sets Homework Problem? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
Mathematics news on Phys.org
Hello JPARK, $$|A|\leq |B|\Leftrightarrow \exists f:A\to B\mbox{ injective}\\
|B|\leq |C|\Leftrightarrow \exists g:B\to C\mbox{ injective}$$ But $g\circ f:A\to C$ is injective (the composition of injective maps is injective), hence $|A|\leq |C|$.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top