# Understanding these basic definition and statement~

• If $x$ belongs to a class $A$, we write $x \in A$ and read as ” $x$ is an element of $A . “$ Otherwise, we write $x \notin A$.
• If $A$ and $B$ are sets, then $A \subseteq B(” A$ is a subset of $B$ ” or “A is contained in $B “$ ) means that each element of $A$ is also an element of $B$. Sometimes we write $B \supseteq A$ (“B contains A”) instead of $A \subseteq B$.
• We say two sets $A$ and $B$ are equal, written $A=B$, if $A \subseteq B$ and $B \subseteq A$.
• Any statement $S$ has a negation $\sim S($ “not $S$ “) defined by
$\sim S$ is true if $S$ is false and $\sim S$ is false if $S$ is true.
• Let $P(x)$ denote a property $P$ of the object $x$. We write $\exists$ for the quantifier “there exists.” The expression
$$\exists x \in X: P(x)$$
means that “there exists (at least) one object $x$ in the class $X$ which has the property $P . “$ The symbol $\exists$ is called the existential quantifier.
• We use the symbol $\forall$ for the quantifier “for all.” The expression
$$\forall x \in X: P(x)$$
has the meaning “for each object $x$ in the class $X, x$ has property $P . “$ The symbol $\forall$ is called the universal quantifier (or sometimes the general quantifier).
• We use the symbol := to mean “is defined by.” We take $x:=y$ to mean that the object or symbol $x$ is defined by the expression $y$.
• Note that for negation of a statement we have:
(i) $\sim \sim A:=\sim(\sim A)=A$
(ii) $\sim(A$ and $B)=(\sim A)$ or $(\sim B)$
(iii) $\sim(A$ or $B)=(\sim A)$ and $(\sim B)$
(iv) $\sim(\forall x \in X: P(x))=(\exists x \in X: \sim P(x))$
(v) $\sim(\exists x \in X: P(x))=(\forall x \in X: \sim P(x))$.
• Let $A$ and $B$ be statements. $A$ implies $B$ will be denoted by $A \Rightarrow B$. If $A$ implies $B$, we take this to mean that if we wish to prove $B$, it suffices to prove $A(A$ is a sufficient condition for
B).
• The equivalence $A \Leftrightarrow B$ (“A and $B$ are equivalent” or ” $A$ if and only if $B$,” often written $A$ iff $B$ ) of the statements $A$ and $B$ is defined by
$$(A \Leftrightarrow B):=(A \Rightarrow B) \text { and }(B \Rightarrow A)$$
$A$ is a necessary and sufficient condition for $B$, or vice versa.
• The statement $\sim B \Rightarrow \sim A$ is called the contrapositive of the statement $A \Rightarrow B$. In standard logic practices, any statement is considered equivalent to its contrapositive. It is often easier to prove a statement’s contrapositive instead of directly proving the statement itself.
• To prove $A \Rightarrow B$ by contradiction, one supposes $B$ is false (that $\sim B$ is true). Then, also assuming that $A$ is true, one reaches a conclusion $C$ which is already known to be false. This contradiction shows that if $A$ is true $\sim B$ cannot be true, and hence $B$ is true if $A$ is true.
• Given two sets $A$ and $B$, we define $A \cup B$ (“the union of $A$ with $B$ “) as the set
$$A \cup B:={x: x \in A \text { or } x \in B \text { or both }}$$
When speaking about unions, if we say $x \in A$ or $x \in B$ it also includes the possibility that $x$ is in both $A$ and $B$.
• We define $A \cap B$ (“the intersection of $A$ with $B$ “) as the set
$$A \cap B:={x: x \in A \text { and } x \in B} \text {. }$$
• Let $A$ and $B$ be subsets of $X .$ Then
$$A \backslash B:={x \in X: x \in A \text { and } x \notin B}$$
is the relative complement of $B$ in $A$. When the set $X$ is clear from the context we write also
$$A^{c}:=X \backslash A$$
and call $A^{c}$ the complement of $A$.
• If $X$ is a set, then so is its power set $\mathcal{P}(X)$. The elements of $\mathcal{P}(X)$ are the subsets of $X$. Sometimes the power set is written $2^{X}$ for a reason which is made clear in Problem $2.8$.
• Let $f: X \rightarrow Y$ be a function, then
$$i m(f):={y \in Y ; \exists x \in X: y=f(x)}$$
is called the image of $f$. We say $f$ is surjective (or onto) if $i m(f)=Y$, injective (or one-to-one) if $f(x)=f(y)$ implies $x=y$ for all $x, y \in X$, and $f$ is bijective if $f$ is both injective and surjective.
• If $X$ and $Y$ are sets, the Cartesian product $X \times Y$ of $X$ and $Y$ is the set of all ordered pairs $(x, y)$ with $x \in X$ and $y \in Y$.
• Let $X$ be a set and $\mathcal{A}=\left{A_{i}: i \in I\right}$ be a family of sets and $I$ is an index set. Intersection and union of this family are given by
$$\bigcap_{i \in I} A_{i}=\left{x \in X ; \forall i \in I: x \in A_{i}\right}$$
and
$$\bigcup_{i \in I} A_{i}=\left{x \in X ; \exists i \in I: x \in A_{i}\right}$$
• Inverse image of $B$ under $f$ (or pre-image of $B$ ), $f^{-1}(B)$ defined as
$$f^{-1}(B)={x \in X: f(x) \in B} .$$
Note that we can form $f^{-1}(B)$ for a set $B \subset Y$ even though $f$ might not be one-to-one or onto.
• We will use standard notation, $\mathbb{N}$ for the set natural numbers, $\mathbb{Z}$ for the set of integers, $\mathbb{Q}$ for the set rational numbers, and $\mathbb{R}$ for the set real numbers. We have the natural containments:
$$\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R}$$
• Two sets $A$ and $B$ have the same cardinality if there is a bijection from $A$ to $B$. In this case we write $A \sim B$. We say $A$ is countable if $\mathbb{N} \sim A$. An infinite set that is not countable is called an uncountable set.

# Try the following problem to test yourself！

Problem 1.

Problem 1.1 Consider the four statements
$$\begin{array}{ll} \text { (a) } & \exists x \in \mathbb{R} \forall y \in \mathbb{R} & x+y>0 ; \ \text { (b) } & \forall x \in \mathbb{R} \exists y \in \mathbb{R} & x+y>0 ; \end{array}$$
(c) $\forall x \in \mathbb{R} \forall y \in \mathbb{R} \quad x+y>0$
(d) $\quad \exists x \in \mathbb{R} \forall y \in \mathbb{R} \quad y^{2}>x$.

1. Are the statements $a, b, c, d$ true or false?
2. Find their negations.

Proof .

faf

1. (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 $x+y=x-x-1=-1 \leq 0 .$
2. (b) is true. Indeed for $x \in \mathbb{R}$, one can take $y=-x+1$ which gives $x+y=1>0$. The negation of $(\mathrm{b})$ is $\exists x \in \mathbb{R} \forall y \in \mathbb{R} \quad x+y \leq 0$.
3. $(\mathrm{c}): \forall x \in \mathbb{R} \forall y \in \mathbb{R} \quad x+y>0$ is false. Indeed one may take $x=-1, y=0$. The negation of (c) is $\exists x \in \mathbb{R} \exists y \in \mathbb{R} x+y \leq 0$.
4. (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.

Problem $1.2$ Let $f: \mathbb{R} \rightarrow \mathbb{R}$. Find the negations of the following statements:

1. For any $x \in \mathbb{R} f(x) \leq 1$.
2. The function $f$ is increasing.
3. The function $f$ is increasing and positive.
4. There exists $x \in \mathbb{R}^{+}$such that $f(x) \leq 0$.
5. There exists $x \in \mathbb{R}$ such that for any $y \in \mathbb{R}$, if $xf(y)$.

Proof .

faf

1. This statement may be rewritten as: (For every $x \in \mathbb{R})(f(x) \leq 1)$. The negation of “( For every $x \in \mathbb{R})$ ” is “There exists $x \in \mathbb{R}$ ” and the negation of ” $(f(x) \leq 1)$ ” is ” $f(x)>1 . “$ Hence the negation of the statement is: “There exists $x \in \mathbb{R}, f(x)>1 . “$
2. First let us rewrite the statement “The function $f$ is increasing”: “for any real numbers $\left(x_{1}, x_{2}\right)$, if $x_{1} \leq x_{2}$ then $f\left(x_{1}\right) \leq f\left(x_{2}\right) . “$ This may be rewritten as: “(for any real numbers $x_{1}$ and $\left.x_{2}\right) \quad\left(x_{1} \leq x_{2}\right.$ implies $\left.f\left(x_{1}\right) \leq f\left(x_{2}\right)\right) . ” \quad$ The negation of the first part is: “(there exists a pair of real numbers $\left.\left(x_{1}, x_{2}\right)\right)$ ” and the negation of the second part is: “( $x_{1} \leq x_{2}$ and $\left.f\left(x_{1}\right)>f\left(x_{2}\right)\right)$ “. Hence the negation of the complete statement is: “There exist $x_{1} \in \mathbb{R}$ and $x_{2} \in \mathbb{R}$ such that $x_{1} \leq x_{2}$ and $f\left(x_{1}\right)>f\left(x_{2}\right) .$ ‘
3. The negation is: the function $f$ is not increasing or is not positive. We already did describe the statement “the function $f$ is not increasing.” Let us focus on “the function $f$ is not positive.” We get: “there exists $x \in \mathbb{R}, f(x)<0 . “$ Therefore the negation of the complete statement is: “there exist $x_{1} \in \mathbb{R}$ and $x_{2} \in \mathbb{R}$ such that $x_{1}<x_{2}$ and $f\left(x_{1}\right) \geq f\left(x_{2}\right)$, or there exists $x \in \mathbb{R}, f(x)<0 . “$
4. This statement may be rewritten as follows: “(there exists $\left.x \in \mathbb{R}^{+}\right)(f(x) \leq 0) . “$ The negation of the first part is: “(for any $\left.x \in \mathbb{R}^{+}\right), “$ and for the second part:” $(f(x)>0) . “$ Hence the negation of the complete statement is: “for any $x \in \mathbb{R}^{+}, f(x)>0 . “$
5. This statement may be rewritten as follows: ” $(\exists x \in \mathbb{R})(\forall y \in \mathbb{R})(xf(y)) . “$
The negation of the first part is: ” $(\forall x \in \mathbb{R}), “$ for the second part: ” $(\exists y \in \mathbb{R}), “$ and for the third part: ” $(x<y$ and $f(x) \leq f(y)) . “$ Hence the negation of the complete statement is:
” $\forall x \in \mathbb{R}, \exists y \in \mathbb{R}, x<y$ and $f(x) \leq f(y) . “$

Fourier analysis代写

## 离散数学代写

Partial Differential Equations代写可以参考一份偏微分方程midterm答案解析