Problem 1 Consider the four statements
(a) $\exists x \in \mathbb{R} \forall y \in \mathbb{R} \quad x+y>0$;
(b) $\forall x \in \mathbb{R} \quad \exists y \in \mathbb{R} \quad x+y>0$;
(c) $\forall x \in \mathbb{R} \forall y \in \mathbb{R} \quad x+y>0$;
(d) $\exists x \in \mathbb{R} \forall y \in \mathbb{R} \quad y^2>x$.
1.Are the statements $a, b, c, d$ true or false?
2.Find their negations.
(a) is false. Since its negation $\forall x \in \mathbb{R} \exists y \in \mathbb{R} \quad x+y \leq 0$ is true. Because if $x \in \mathbb{R}$, there exists $y \in \mathbb{R}$ such that $x+y \leq 0$. For example, we may take $y=-(x+1)$ which gives $x+y=x-x-1=-1 \leq 0$.
(b) is true. Indeed for $x \in \mathbb{R}$, one can take $y=-x+1$ which gives $x+y=1>0$. The negation of (b) is $\exists x \in \mathbb{R} \forall y \in \mathbb{R} \quad x+y \leq 0$.
(c) : $\forall x \in \mathbb{R} \forall y \in \mathbb{R} \quad x+y>0$ is false. Indeed one may take $x=-1, y=0$. The negation of (c) is $\exists x \in \mathbb{R} \exists y \in \mathbb{R} x+y \leq 0$.
(d) is true. Indeed one may take $x=-1$. The negation is: $\forall x \in \mathbb{R} \exists y \in \mathbb{R} \quad y^2 \leq x$.
Problem 2 Find the negation of the following statements:
- Any rectangular triangle has a right angle.
- In all the stables, the horses are black.
- For any integer $x \in \mathbb{Z}$, there exists an integer $y \in \mathbb{Z}$ such that, for any $z \in \mathbb{Z}$, the inequality $z<x$ implies $z<x+1$.
- $\forall \varepsilon>0 \exists \alpha>0 /|x-7 / 5|<\alpha \Rightarrow|5 x-7|<\varepsilon$.
- A triangle with no right angle, is not rectangular.
- There exists a stable in which there exists at least one horse who is not black.
- If we rewrite the statement in mathematical language:
$$
\forall x \in \mathbb{Z} \quad \exists y \in \mathbb{Z} \quad \forall z \in \mathbb{Z} \quad(z<x \quad \Leftrightarrow \quad z<x+1),
$$
the negation is
$$
\exists x \in \mathbb{Z} \quad \forall y \in \mathbb{Z} \quad \exists z \in \mathbb{Z} \quad(z<x \text { and } z \geq x+1) .
$$ - $\exists \varepsilon>0 \quad \forall \alpha>0 \quad(|x-7 / 5|<\alpha$ and $|5 x-7| \geq \varepsilon)$.
Problem 3 For two sets $A$ and $B$ show that the following statements are equivalent:
a) $A \subseteq B$
b) $A \cup B=B$
c) $A \cap B=A$
$$
(a \Rightarrow b)
$$
Suppose $A \subseteq B$. Let $x \in A \cup B$, then $x \in A$ or $x \in B$. If $x \in A$, then since $A \subseteq B$, we have $x \in B$. Thus, for any $x \in A \cup B, x \in B$, so $A \cup B \subseteq B$. Let $x \in B$, then $x \in A \cup B$ so $B \subseteq A \cup B$, and hence $A \cup B=B$.
$$
(b \Rightarrow c)
$$
Let $x \in A \cap B \Rightarrow x \in A$ so $A \cap B \subseteq A$. Let $x \in A \Rightarrow x \in A \cup B=B$ so $x \in B$ and therefore $x \in A \cap B$ so $A \subseteq A \cap B$. Thus $A \cap B=A$.
$$
(c \Rightarrow a)
$$
Let $x \in A$, then by hypothesis $x \in A \cap B$, which in turn implies that $x \in B$ as well. Thus $A \subseteq B$
Problem 4 If $A$ and $B$ are sets, then show that
a) $\mathcal{P}(A) \cup \mathcal{P}(B) \subseteq \mathcal{P}(A \cup B)$
b) $\mathcal{P}(A) \cap \mathcal{P}(B)=\mathcal{P}(A \cap B)$
a)
$$
\begin{aligned}
X \in \mathcal{P}(A) \cup \mathcal{P}(B) & \Rightarrow X \subseteq A \text { or } X \subseteq B \
& \Rightarrow X \subseteq A \cup B \
& \Rightarrow X \subseteq \mathcal{P}(A \cup B) .
\end{aligned}
$$
b)
$$
\begin{aligned}
X \in \mathcal{P}(A) \cap \mathcal{P}(B) & \Leftrightarrow & X \subseteq A \text { and } X \subseteq B \
& \Leftrightarrow & X \subseteq A \cap B \
& \Leftrightarrow & X \in \mathcal{P}(A \cap B) .
\end{aligned}
$$
Problem 5 For an arbitrary function $f: X \longrightarrow Y$, prove the following identities:
a) $f^{-1}\left(\bigcup_{i \in I} B_i\right)=\bigcup_{i \in I} f^{-1}\left(B_i\right)$.
b) $f^{-1}\left(\bigcap_{i \in I} B_i\right)=\bigcap_{i \in I} f^{-1}\left(B_i\right)$.
c) $f^{-1}\left(B^c\right)=\left[f^{-1}(B)\right]^c$.
a)
$$
\begin{aligned}
x \in f^{-1}\left(\bigcup_{i \in I} B_i\right) & \Leftrightarrow \quad f(x) \in \bigcup_{i \in I} B_i \
& \Leftrightarrow \quad \exists i \in I \text { such that } f(x) \in B_i \
& \Leftrightarrow \quad \exists i \in I \text { such that } x \in f^{-1}\left(B_i\right) \
& \Leftrightarrow \quad x \in \bigcup_{i \in I} f^{-1}\left(B_i\right) .
\end{aligned}
$$
b)
$$
\begin{aligned}
& x \in f^{-1}\left(\bigcap_{i \in I} B_i\right) \Leftrightarrow \quad f(x) \in \bigcap_{i \in I} B_i \
& \Leftrightarrow \quad f(x) \in B_i \quad \forall i \in I \
& \Leftrightarrow \quad x \in f^{-1}\left(B_i\right) \quad \forall i \in I \
& \Leftrightarrow \quad x \in \bigcap_{i \in I} f^{-1}\left(B_i\right) \text {. } \
&
\end{aligned}
$$
c)
$$
\begin{aligned}
x \in f^{-1}\left(B^c\right) & \Leftrightarrow & f(x) \in B^c \
& \Leftrightarrow & f(x) \notin B \
& \Leftrightarrow & x \notin f^{-1}(B) \
& \Leftrightarrow & x \in\left[f^{-1}(B)\right]^c .
\end{aligned}
$$
