Suppose $A, B, C$ are sets. Show that if $|A| \leq|B|$ and $|B| \leq|C|$, then $|A| \leq|C|$.

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∻Solution by Steps∻ ‖ step 1 ⋮ To show that if $|A| \leq |B|$ and $|B| \leq |C|$, then $|A| \leq |C|$, we can use the transitive property of inequalities. ‖ ‖ step 2 ⋮ Since $|A| \leq |B|$, there exists an injection from set $A$ to set $B$. Similarly, since $|B| \leq |C|$, there exists an injection from set $B$ to set $C$. ‖ ‖ step 3 ⋮ By composing these injections, we can construct an injection from set $A$ to set $C$, which implies that $|A| \leq |C|$. ‖ ∻Answer∻ ⚹ $|A| \leq |C|$ ⚹ ∻Key Concept∻ ⚹Transitive property of set sizes (cardinalities)⚹ ∻Explanation∻ ⚹The transitive property allows us to conclude that if one set can be injected into another, and that set can be injected into a third, then the first set can also be injected into the third, establishing the inequality of their sizes.⚹◊What is the relationship between the cardinality of sets and the concept of injections in set theory?⍭ Generate me a similar question◊

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