Table of Contents

Problems
Problem 1 (Herstein Problem)
Problem:If p is a prime >3, and $1+\frac{1}{2}+\cdots+\frac{1}{(p1)}=a/b$, then prove that $p^2a$.
Solution:
We need to consider the result : if $i<p$ , then
(1)Now, note that$(a/bp)=(a/p)\cdot(1/b)$; but b isn't divisible by p and pa from well known result, hence as $frac{a}{p}$ and b are integers, with b coprime to p, we have that $[(a/bp)]_p\in \mathbb Z_p$. Define $x:=(p1)/2$.
(2)so that, twice the left hand side becomes
(3)The justification of the last equality to zero : 3 divides either of (p1) and (p+1), and in the latter case, $2(p1)$ and $3\{2(p+1)3\}=(2p1)$, whence $6(p1)(2p1)$ always.
As 2 and p are coprime, we can see that $0=[a/p]_p\cdot[b]^{1}_p$ so that $[a/p]_p=0$. Hence $p^2a$.
Problem 2 (cardinality of bijections)
Let an infinite set A be of cardinality $\alpha$.
We want to determine the cardinality of the set of all bijections :$A\leftrightarrow A$.
Obviously, it is less than or equal to $card(A^A)\;=\;\alpha^\alpha\le 2^\alpha$.
We'd show that there are at least $2^\alpha$ such bijections.
THEOREM.The cardinality of the sets of all oneone, onto functions : $A\to A$ is $2^\alpha$.
Call a subset $S$ of $A$ fair iff $card(A)=card(S)=card(S\backslash A)$. Before proving the theorem, we have the following lemma:
LEMMA.The cardinality of the set of all the fair subsets of $A$ is $2^\alpha$.
proof.
First, the cardinality of that set is evidently $\le\alpha^\alpha=2^\alpha$.
We have, $A\times A\sim A$. So, if $f:A\times A\leftrightarrow A$, and $S$ be a fair subset of $A\times A$, then $f(S)$ is a fair subset of $A$; this is easy.
Now, for each nonvoid $K\subsetneqq A$, the set $K\times A$ is a fair subset of $A\times A$. But there are $2^\alpha$ such subset $K$ of $A$. So, we have $2^\alpha$ fair subsets of $A\times A$, and so, exactly that many fair subsets of $A$.
Now we prove our main theorem:
proof.
Fix a fair subset $K$ of $A$.
For each fair subset $X$ of $A$, we have a function $f_X:A\to A$ defined as follows:
$f_X\upharpoonright X=\varphi$, where $\varphi$ is the bijection :$X\leftrightarrow K$;
$f_X\upharpoonright(A\backslash X)=\psi$, where $\psi$ is the bijection :$(A\backslash X)\leftrightarrow K^c$.
Clearly, $f_X$ is a bijection:$A\leftrightarrow A$.
See that $f_X^{\leftarrow}(K)=X$. Hence, for two distinct fair subsets of $A$, we would get two distinct such bijections.
What we have proved is that there are at least as many bijections as the fair subsets, and so, there are, by the lemma, at least $2^\alpha$ such bijections; and as there are at most $2^\alpha$ bijections, by Schröeder Bernstein, there are precisely $2^\alpha$ bijections : $A\leftrightarrow A$.
Problem 3
We need two propositions:
 A linear ordered space $(S,<,\mathcal T_<)$ is compact iff it is a complete lattice (i.e., every subset has a lub in S).
 It is connected iff it is dense and Dedekind complete.
In view of the above, we have the following result:
Given linear ordered space S, and a sequence of nonempty compact subsets $\{A_n:n\in\mathbb N\}$ in such a way that $A_{n+1}\subseteq A_n$, we must have $\bigcap_{n\in\mathbb N}A_n \ne\emptyset$.
This is pretty simple. As any linear order is Hausdorff, any compact subset is closed; and the family $\{A_n:n\in\mathbb N\}$ of nonempty closed sets has finite_intersection_property, and each is a subset of the compact set $A_1$; so, the intersection must be nonempty.
The theorem may also have been stated, when S is connected and by 2, dense and complete, with "compact subsets" replaced by "closed bounded sets", because, such a set is, then, a complete lattice, and by 1, is compact:
Given connected linear order S, and a sequence of nonempty closed, bounded sets $\{A_n:n\in\mathbb N\}$ in such a way that $A_{n+1}\subseteq A_n$, we must have $\bigcap_{n\in\mathbb N}A_n \ne\emptyset$.
Problem 4 (Baire's Theorem)
The axiom_of_dependent_choice looks like:
AXIOM (DC). If R is a binary relation on a nonempty set X such that for all a in X, there is b in X with aRb, then there is a sequence $\{x_n:n\in\omega\}$ with the property that $\forall n\in\omega\colon x_n\;R\;x_{n+1}$.
From that we prove the following statement
THEOREM (Baire). If X be a locally compact, Hausdorff_space space, and $\{G_n:n\in\omega\}$ be a set of dense, open subsets, then $\bigcap\{G_n:n\in\omega\}$ is dense.
proof. Consider any nonempty open set V. We have to show that $V\cap(\bigcap\{G_n:n\in\omega\})\ne\emptyset$.
Putting $A_0=V$, consider the set
(4)Call it Y. Evidently, Y is not empty since X is locally compact and G are dense, open. We impose a relation R on Y and define R by
$(A_1,\cdots,A_n)\;R\;(B_1,\cdots,B_m)$ if and only if
 m= n+1
 for each $i\le n,\;A_i=B_i$.
See that, as X is regular and locally compact, for each $(A_1,\cdots,A_n)$ we can get $(A_1,\cdots,A_n,A_{n+1})$ with $(A_1,\cdots,A_n)\;R\;(A_1,\cdots,A_n,A_{n+1})$. So, the structure (Y,R) satisfies the requirement of DC.
So, there is a sequence $\{z_k:k\in\omega\}$ with $z_k\;R\;z_{k+1}$.
Suppose that $z_o=(A_1,\cdots,A_m)$; then for each k>0, $z_k=(A_1,\cdots,A_m,\cdots,A_{m+k})$.
We see that these $A_i$ are compact subsets of V. Moreover, $\forall i\in\mathbb N\colon A_i\subseteq G_i$.
Now, as X is Hausdorff, each compact set is closed. The family $\{A_i:i\in\mathbb N\}$ is a family of nonempty closed subsets of the compact set $A_1$, with finite_intersection_property. Hence the intersection $\bigcap\{A_i:i\in\mathbb N\}$ is nonempty.
As for each i, $A_i\subseteq G_i\cap V$, we have that
$\emptyset\ne\bigcap\{A_i:i\in\mathbb N\}\subseteq V\cap(\bigcap\{G_n:n\in\omega\})$.
So we have proved the theorem.
What we learn is that the intersection of a countable dense sets is dense, which is to say that locally compact, Hausdorff spaces are Baire_space.
The method in which we developed the whole argument upon DC, is used in many proofs in [2].
In the case when X is a complete metrizable space, the proof goes in much the same way, with a meagre modification.
Problem 5 (Hartog's Number)
Existence of Hartogs's number
If possible, suppose that X is a nonempty set and for any well ordered set $\alpha$ there is an injective mapping from $\alpha$ to X.
Consider the set $A_0:=\{(W,\le)\colon W\subseteq X;\;\le\mbox{ well orders }W\}$, which is a set, for it may be identified with a subset of $\wp(X)\times \wp(X\times X)$, since a well order on $W$ is actually a subset of $X\times X$.
As each well ordered set is orderisomorphic with a unique ordinal by a unique order isomorphism, $A_0$ gives rise via Replacement to the set $A:=\{\alpha\in\mbox{On}\colon (\exists W\in A_0)(\alpha\cong W)\}$.
We'd show that A is an ordinal. As members of A are ordinals, we only have to show that A is $\in$transitive. To see that, consider $\beta\in\alpha;\;\alpha\in A$. But there is W, well ordered subset of X, and the order isomorphism $f$ that identifies $\alpha$ with W. Under this order isomorphism, the image of $\beta$ is in $A_0$; so that (under the restriction map $f\upharpoonright\beta$) the ordinal $\beta\in A$, proving that A is $\in$transitive.
Now there is a oneone map $g\colon A\rightarrowtail X$ and as A is well ordered, g(A) is a well ordered subset of X. So, A is an ordinal, which is order isomorphic with the well ordered subset g(A) of X. This means that A is in A, a contradiction.
Problem 6
Noetherian rings:
Counterexample to the following statement:
Let $R$ be a ring and $N$ be its nilradical. If N is nilpotent and R/N is noetherian, then R is also noetherian.
Let $R$ be the quotient of $\mathbb Z[X_1,X_2,\cdots]$ by the ideal $\langle X_1,X_2,\cdots\rangle^2$; and see that its nilradical consists of polynomials which have no constant terms. Surely, the square of the nilradical is zero, whence it is nilpotent. But $R/N$ is nothing but $\mathbb Z$, whence is noetherian. One can easily check that the nilradical is not finitely generated, so that $R$ is not noetherian itself.
Problem 7 (abelian)
Theorem. If G is a group and there are two coprime naturals m and n with $(\forall a,b\in G)(a^nb^n=b^na^n;\;a^mb^m=b^ma^m)$ then the group is abelian.
Proof. Now we have $r,s\in\mathbb Z$ with $mr+ns=1$. We first propose to show that
(5)Case 1. mr>0.
(6)Case 2. mr<0.
Let $mr=p$, $p\in\mathbb N$.
Then the second step:
(8)Similarly,
(9)Now the final step: for any x,y in G,
(11)Problem 8 (A nonlinear Lie group)
Let $G$ be the universal cover of the Lie group $SL_2(\mathbb R)$. We have the complexification ${\mathfrak sl}_2(\mathfrak C)$ of the lie algebra ${\mathfrak sl}_2(\mathbb R)$, and as $G$ is simply connected, the lie algebra inclusion ${\mathfrak sl}_2(\mathbb R)\hookrightarrow {\mathfrak sl}_2(\mathbb C)$ lifts to a morphism of lie groups $G\to SL_2(\mathbb C)$. Suppose $G\to GL_n(\mathbb R)$ is a representation of $G$; then as $SL_2(\mathbb C)$ is simply connected, it lifts to a lie group morphism $SL_2(\mathbb C)\to GL_n(\mathbb C)$. We have the commutative diagram
This diagram shows that the image accross the lower horizontal line of $SL_2(\mathbb R)$ lies actually inside $GL_n(\mathbb R)$ (by lifting an element to $G$ etc.), which means that the representation of $G$ factors through $SL_2(\mathbb R)$ i.e. is not injective. Thus, $\widetilde{SL_2(\mathbb R)}$ cannot be linear.
Problem 9 (Extended Lagrange's Theorem)
Theorem. If H is a subgroup of a group G, and if we denote by (G:H) the set of all left cosets of H in G, then
$(G:H)\times H\;$ has the same cardinality as G.
Proof. We would rely upon axiom of choice. (G:H) is a set of distinct elemtens. Let us have a choice function $ch:(G:H)\to G$ that assigns to each left coset $\alpha$ of H in G with a unique element $ch(\alpha)=g_\alpha\in G$ so that $g_\alpha H=\alpha$.
Now, note that for each $g\in G$ there is a unique coset $\alpha$ containing $g$, so there is a unique $g_\alpha\in G$ with $g\in\alpha=g_\alpha H$. Not only that, but we also have $h\in H$ with $g=g_\alpha\cdot h$. The assignment $g\mapsto (g_\alpha H,h)$ is well defined since the corresponding left coset is unique, and once it is fixed we can easily see the uniqueness of $h$. We name the assignment $f$
This map $f$ is, moreover, oneone since given $f(g)=(ch(\alpha)H,h)=(ch(\beta)H,h')=f(g')$, we must have that $\alpha=\beta,\; h=h'$, so that $g=ch(\alpha)\cdot h=ch(\beta)\cdot h'=g'$.
That it is onto is clear.
Problem 10(Baire Spaces etc.)
Let $X$ be a topological space. If the subsets $E$ and $E^c$ are both dense, and $E$ a $G_\delta$ and an $F_\sigma$, then
(12)for open $G$'s and closed $F$'s; and as $E$ is dense, so is each $G_n$, and as $E^c=\cap_n F_n^c$ is dense, so is each $F_n^c$.
Now consider the family $\{G_n:n\in\mathbb N\}\cup\{F_n^c:n\in\mathbb N\}$, which consists simply of a countable number of dense, open sets.
If the space concerned is a Baire space, then the intersection of the said family would be dense; but in reality it is $E\cap E^c=\emptyset$.
So, what we get:
Theorem 1. In a Baire space, if a subset $E$ is dense with its complement $E^c$ also dense, at most one of $E$ and $E^c$ can be $G_\delta$.
Corollary 2.
$\mathbb Q$ is not a $G_\delta$, since it is an $F_\sigma$ as $\mathbb Q$ being countable can be expressed as a union of countable number of singletons and singletons are nowhere dense.
Now consider a continuous function $f:X\to \mathbb R$, where $X$ is a complete or locally compact metric space. Given $x\in X$, consider $\omega(x):=\inf_{\epsilon>0}\delta(f(B(x,\epsilon)))$, which is an extended real number. Consider any positive $\eta$, and Suppose that $\omega(x_0)<\eta$, which means that there is some $\epsilon>0$ such that $\delta(f(B(x_0,\epsilon)))<\eta)$.
We claim that $\omega( B(x_0,\epsilon))<\eta$. For, given such an $x$, define $\epsilon'=\min\{d(x,x_0),\epsilond(x,x_0)\}$, and $B(x,\epsilon')\subseteq B(x_0,\epsilon)$, so that $\delta(B(x,\epsilon'))\le\delta(B(x_0,\epsilon))<\eta$ and it means that $\omega(x)<\eta$. What we have proved is actually that, if at some point $x_0$ the $\omega$ function is less than $\eta$, then there is an open neighbourhood of $x_0$ in which $\omega$ is still less than $\eta$, in other words, the set $\{x:\omega(x)<\eta\}$ is open for any positive $\eta$. So,
Lemma 3. $\{x:\omega(x)<\eta\}$ is open for any positive $\eta$.
Observation:
$\omega:X\to \mathbb R$ is an upper semicontinuous function.
Now observe that, $f$ is continuous at $x$ iff $\omega(x)=0$. Let us call $G_n=\{x:\omega(x)<1/n\}$, then
(13)And it immediately follows that
Theorem 4. Given a function $f:X\to \mathbb R$, where X is a complete or locally compact metric space, then the points at which $f$ is continuous constitute a $G_\delta$.
Corollary 5.
There is no function $f:\mathbb R\to \mathbb R$ which is continuous on the rationals, and discontinuous on the irrationals.
Remark
Now, in the previous discussion, if instead we take $(X,\mathcal T)$ to be any Baire space, and define $\omega(x):=\inf\{\delta(f(V)):x\in V\in\mathscr T\}$ which is again an extended real number, then we can say that $f$ is continuous at $x$ iff $\omega(x)=0$. In that case, if $\omega(x_0)<\eta$, then there is some $V\in\mathcal T$ with $\delta(f(V))<\eta$, and that $V$ works for all $x\in V$, so that Lemma 3. holds true. And as well, Equation (13) holds, so that we can extend Theorem 4. to any function $f:X\to \mathbb R$, where $X$ is any Baire space.