MHB JPARK 's question at Yahoo Answers (Cardinality)

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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.
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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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