Answer
The mistake is in assuming that $A \subseteq B$ implies there exists an $x\in A$. This is invalid, especially if $ A = \emptyset$.
Work Step by Step
This “proof” is **incorrect**, and the mistake lies in its **misuse of logic about subset definitions**.
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## ✅ What the proof is trying to show:
For all sets \(A\), \(B\), and \(C\):
If \(A \subseteq B\) and \(B \subseteq C\), then \(A \subseteq C\).
This statement is **true**, and it's a standard transitive property of subsets.
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## ❌ The Error in the “Proof”
The incorrect proof says:
> “Since \(A \subseteq B\), there is an element \(x\) such that \(x \in A\) and \(x \in B\)…”
This is **false logic**.
Just because \(A \subseteq B\), that doesn't mean **there exists an \(x \in A\)** — for example, what if \(A = \emptyset\)?
The subset relation does **not** imply any particular element exists.
Similarly, the later statement:
> “Since \(B \subseteq C\), there is an element \(x\) such that \(x \in B\) and \(x \in C\)”
is also incorrectly assuming the existence of elements.
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## ✅ The Correct Reasoning
To prove \(A \subseteq C\), the correct method is:
> Let \(x \in A\) (assume arbitrary element).
> Since \(A \subseteq B\), we know \(x \in B\).
> Since \(B \subseteq C\), we then get \(x \in C\).
> Therefore, \(x \in C\), so \(A \subseteq C\).
That is, you **start with an arbitrary element of \(A\)** and use the subset properties to show it must be in \(C\).
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## 🧠 Summary of the Mistake
The flawed proof incorrectly argues:
> “\(A \subseteq B \Rightarrow\) some element exists in \(A\)” — ❌ not necessarily true.
The correct logic would be:
> “**For all** \(x \in A\), \(x \in B\), and then \(x \in C\)” — ✅ universal reasoning, not existential.
