Answer
(a) \(\exists S\ \forall T,\ S \cap T \ne \emptyset\) ✅ **Original is true**
(b) \(\forall S\ \exists T,\ S \cup T \ne \emptyset\) ✅ **Negation is true**
Work Step by Step
Let’s analyze and **negate** each of the two statements and determine whether the **original** or the **negation** is true.
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## 🔹 **(a)** Statement:
> \(\forall\) sets \(S\), \(\exists\) a set \(T\) such that \(S \cap T = \emptyset\)
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### ✅ **Negation**:
Negating \(\forall S\ \exists T \ (S \cap T = \emptyset)\) gives:
> \(\exists\) a set \(S\) such that \(\forall\) sets \(T\), \(S \cap T \neq \emptyset\)
In plain English:
“There is some set \(S\) that intersects **every** set \(T\) nontrivially.”
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### ❌ Is the **original** statement true?
Yes!
For **any** set \(S\), we can always find a set \(T\) that has **no overlap**.
For example:
- Let \(T = \emptyset\), then \(S \cap T = \emptyset\)
- Or let \(T\) be any set that contains elements **not in** \(S\)
✅ So the **original** is **true**.
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### ✅ Final answer for (a):
- **Negation**:
\(\exists S\ \forall T,\ S \cap T \neq \emptyset\)
- ✅ The **original** is **true**, the negation is **false**.
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## 🔹 **(b)** Statement:
> \(\exists\) a set \(S\) such that \(\forall\) sets \(T\), \(S \cup T = \emptyset\)
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### ✅ **Negation**:
Negating \(\exists S\ \forall T \ (S \cup T = \emptyset)\) gives:
> \(\forall\) sets \(S\), \(\exists\) a set \(T\) such that \(S \cup T \ne \emptyset\)
In plain English:
“For **every** set \(S\), we can find some \(T\) such that their union is **not** empty.”
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### ❌ Is the **original** statement true?
No.
Let’s suppose the original is true:
> There exists a set \(S\) such that for **every** set \(T\), \(S \cup T = \emptyset\)
But the only way \(S \cup T = \emptyset\) is if **both** \(S = \emptyset\) and \(T = \emptyset\)
But this fails for any nonempty \(T\). So this **cannot** hold for **every** \(T\)
Thus, **no such** \(S\) exists that satisfies the original.
✅ So the **original is false**, and the **negation is true**.
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### ✅ Final answer for (b):
- **Negation**:
\(\forall S\ \exists T\ \text{such that } S \cup T \ne \emptyset\)
- ✅ The **negation** is **true**, original is **false**.
