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 1.1 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) $\quad \exists x \in \mathbb{R} \forall y \in \mathbb{R} \quad y^{2}>x$.
- Are the statements $a, b, c, d$ true or false?
- Find their negations.
faf
- (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 $(\mathrm{b})$ is $\exists x \in \mathbb{R} \forall y \in \mathbb{R} \quad x+y \leq 0$.
- $(\mathrm{c}): \forall x \in \mathbb{R} \forall y \in \mathbb{R} \quad x+y>0$ is false. Indeed one may take $x=-1, y=0$. The negation of (c) is $\exists x \in \mathbb{R} \exists y \in \mathbb{R} x+y \leq 0$.
- (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 $1.2$ Let $f: \mathbb{R} \rightarrow \mathbb{R}$. Find the negations of the following statements:
- For any $x \in \mathbb{R} f(x) \leq 1$.
- The function $f$ is increasing.
- The function $f$ is increasing and positive.
- There exists $x \in \mathbb{R}^{+}$such that $f(x) \leq 0$.
- There exists $x \in \mathbb{R}$ such that for any $y \in \mathbb{R}$, if $xf(y)$.
faf
- 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 $x \in \mathbb{R}, f(x)>1 . “$
- 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 $f\left(x_{1}\right)>f\left(x_{2}\right) .$ ‘
- 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 $x \in \mathbb{R}, f(x)<0 . “$
- 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 $x \in \mathbb{R}^{+}, f(x)>0 . “$
- 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 $f(x) \leq f(y) . “$
黎曼曲面代写,数学代写Riemann surface请认准UprivateTA™. UprivateTA™为您的留学生涯保驾护航。
Fourier analysis代写
微分几何代写
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