JPARK 's question at Yahoo Answers (Cardinality)

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    Cardinality
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SUMMARY

The discussion centers on proving the transitive property of cardinality for sets A, B, and C. It establishes that if the cardinality of set A is less than or equal to that of set B (|A| ≤ |B|) and the cardinality of set B is less than or equal to that of set C (|B| ≤ |C|), then it follows that |A| ≤ |C|. This is demonstrated using the existence of injective functions, specifically that the composition of two injective functions (f: A → B and g: B → C) results in an injective function from A to C (g ∘ f: A → C).

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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.
 
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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 $|A|\leq |C|$.
 

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