The Uniform Cauchy criterion says that $(f_n:E\to \mathbb{R})$ converges uniformly on $E$ if and only ifFor necessity, use the triangle inequality through the uniform limit. For sufficiency, the condition makes $(f_n(x){)}_n$ a Cauchy sequence for each $x$, so completeness of the real numbers defines a pointwise limit $f(x)$. Apply the condition with $\epsilon /2$ and let $m\to \infty$ to obtain $|f_n(x)-f(x)|\le \epsilon /2<\epsilon$ uniformly in $x$.

The sequence of functions $(f_n)$ converges uniformly to $f$ on $E$ ifThe same $N$ must work for every $x$. Equivalently, the supremum norm of $f_n-f$ tends to zero. This is stronger than pointwise convergence, where $N$ may depend on $x$.

Take $y=f(p)$ with $p\in \overline{E}$. Given $\epsilon >0$, continuity gives $\delta >0$ such that $|x-p|<\delta$ implies $|f(x)-f(p)|<\epsilon$. Since $p\in \overline{E}$, choose $x\in E$ with $|x-p|<\delta$. Thus every neighbourhood of $y$ meets $f(E)$, provingEquality need not hold. Let $f(x)=e^x$ and $E=\mathbb{R}$. The exponential function is continuous, butThe additional point zero comes from arguments escaping to $-\infty$, not from a finite point of $\overline{E}$.

The limit of a function $f:E\to \mathbb{R}$ at a limit point $p$ of $E$ is $\ell$ whenThe function converges as $x\to p$ if such a finite $\ell$ exists. The puncture excludes any value assigned at $p$.

For $p\in E$, continuity at $p$ meansWhen $p$ is also a limit point, this says that the limit of a function is $f(p)$. At an isolated point of $E$, every function is continuous relative to $E$, even though the punctured limit does not have a unique forced value.

A point is in the closure $\overline{E}$ exactly when every neighbourhood meets $E$. This is equivalent to belonging to $E$ or being a limit point of $E$.

If $p\notin \overline{E}$, choose $\epsilon >0$ with $(p-\epsilon ,p+\epsilon )\cap E=\varnothing$. Every $q$ with $|q-p|<\epsilon /2$ has a smaller neighbourhood contained in this interval and disjoint from $E$. Thus $(p-\epsilon /2,p+\epsilon /2)$ misses $\overline{E}$, and the complement of $\overline{E}$ is open. Hence $\overline{E}$ is closed.

Every closed set $F$ containing $E$ contains all limit points of $E$: a limit point outside $F$ would have a neighbourhood missing $F$, and hence missing $E$. Therefore $\overline{E}\subseteq F$. Since $\overline{E}$ is itself one of these closed sets,

Suppose $E$ is a closed set and $p\notin E$. Then $p$ is not a limit point of $E$, so some punctured neighbourhood of $p$ misses $E$. As $p\notin E$, the entire neighbourhood misses $E$. Thus $\mathbb{R}\setminus E$ is an open set.

Conversely, suppose every point $p\notin E$ has a neighbourhood disjoint from $E$. No such point can be a limit point of $E$, so every limit point lies in $E$. This proves the equivalence between containing all limit points and having an complement that is an open set.

A limit point $p$ of $E\subseteq \mathbb{R}$ is a point such that, for every $\epsilon >0$, the punctured neighbourhood $(p-\epsilon ,p+\epsilon )\setminus \{p\}$ meets $E$. A closed set contains all of its limit points.

Every nonempty open interval contains a rational number; applying this to $(p,p+\epsilon )$ shows that every $p\in \mathbb{R}$ is a limit point of $\mathbb{Q}$. Thus the limit points of $\mathbb{Q}$ form $\mathbb{R}$, and $\mathbb{Q}$ is not closed.

If $p\in \mathbb{Z}$, radius $1/2$ excludes all other integers. If $p\notin \mathbb{Z}$, its distances to the two neighbouring integers have a positive minimum, and a smaller radius excludes every integer. Thus $\mathbb{Z}$ has no limit points in $\mathbb{R}$ and is closed.

At the nonzero indices $n=m!$ with $m\ge 2$,since $m/(m!)=1/(m-1)!\to 0$. At all other sufficiently large indices the coefficient is zero. The initial coefficients do not affect the limit superior, which equals one. The Cauchy-Hadamard theorem therefore gives $\boxed{R=1}$.

By Stirling's approximation,Here $(2\pi n{)}^{1/(2n)}\to 1$ and $n^{1/n^2}\to 1$, as is seen by taking the natural logarithm. The Cauchy-Hadamard theorem gives $\boxed{R=e^{-1}}$.

The Cauchy-Hadamard theorem states $R^{-1}=\mathrm{limsup}|a_n{|}^{1/n}$. At the prime numbers $n=p$, this root equals $p^{2/p}\to 1$. There are infinitely many prime numbers, so these indices form an unbounded subsequence. For every other positive index the root is zero. The limit superior is therefore one, and $\boxed{R=1}$.

Let $u(x)=(\log x)x^{-2/3}$. Its derivative iswhich is negative for $x>e^{3/2}$. Also $u(x)\to 0$, since a positive power eventually dominates the natural logarithm. Thus the alternating series test proves convergence of $\sum (-1{)}^{n}u(n)$.

For $n\ge 3$, $u(n)\ge n^{-2/3}$, whose p-series diverges. The comparison test therefore rules out absolute convergence. The answer is conditional convergence.

For $t\ge 0$, $|\sin t|\le t$, soThe integral test applies to the decreasing positive comparison function, andThe comparison test proves that the series converges; since all terms are nonnegative, this is also absolute convergence.

The binomial theorem gives $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)\le {\sum }_{k=0}^{2n}\left(\genfrac{}{}{0ex}{}{2n}{k}\right)=4^n$. ThereforeThe geometric series with ratio $4/9$ converges. By the comparison test, the given series is absolutely convergent.

The implication is false. SetThen $n{a}_n=1/\log (n+1)\to 0$. For $n\ge 2$, $\log (n+1)\le \log (2n)\le 2\log n$, so $a_n\ge 1/(2n\log n)$. The integral test givesThus the comparison test shows that $\sum a_n$ diverges.

The implication is true, with absolute convergence. Since $n^{3/2}{a}_n\to 0$, there is $N$ such that $|a_n|\le n^{-3/2}$ for $n\ge N$. The p-series $\sum n^{-3/2}$ converges because its exponent exceeds one. The comparison test proves $\sum |a_n|<\infty$, and absolute convergence implies convergence of $\sum a_n$.

The sine has values $1,0,-1,0$ on the four residue classes of $n$ modulo four. For $n=4k+1$, the exponential function gives $b_n=e^{1-1/n}\to e$; the other residue classes tend to $1,e^{-1},1$. All terms satisfy $b_n\le e$, and the first class approaches $e$. Hence

At even integers $n=2k$, the terms are $1+1/(2k)\to 1$. At odd integers they are $-1-1/(2k-1)\to -1$. The largest eventual values therefore come from the subsequence at even integers, andExplicitly, every sufficiently late term is below $1+\epsilon$, while infinitely many even integers give terms greater than $1-\epsilon$.

For $n\ge 2$ with $n\ne 6$, bijectivity of $\varphi$ and the previous part implyThe supplied numerical fact forces $k=1$. Thus $\varphi$ permutes the transpositions.

Suppose $n\ge 3$, and write $s_r=\varphi ((1r))$ for $r=2,\dots ,n$. These are distinct transpositions. For $r\ne t$, the product $(1r)(1t)$ has order of a group element three, so $s_r{s}_t$ also has order three. Two distinct transpositions have this property precisely when their two-element sets of moved points meet in one point: disjoint ones have a product of order two, whereas ones sharing one point have a three-cycle as product. The two-element support of a permutation associated with each $s_r$ therefore meets that of every other $s_t$.

We now show that they have a common point. Choose two; the support of a permutation is $\{a,b\}$ for one and $\{a,c\}$ for the other, where $a,b,c$ are distinct. Any further two-element set meeting both either contains $a$ or is $\{b,c\}$. If $\{b,c\}$ occurs, any two-element set meeting all three must itself be one of $\{a,b\},\{a,c\},\{b,c\}$: a set containing $a$ must also contain $b$ or $c$, and a set omitting $a$ must contain both $b,c$. Hence a pairwise-intersecting family with no common point consists of exactly those three edges of a triangle.

For $n\ge 5$, there are $n-1\ge 4$ distinct $s_r$, ruling out that triangle. For $n=3$, the two sets already have a common point. For $n=4$, the triangle alternative would make all three $s_r$ fix the fourth point. They would then generate a proper subgroup of $S_4$, contrary to surjectivity of $\varphi$ and part (a)(iv), which imply that the $s_r$ generate $S_4$. Thus in every case $n\ge 3$ there is a common point $a$, and we may writeThe $b_r$ are distinct and different from $a$, so they exhaust the remaining $n-1$ points. Define a permutation $\sigma$ byThenThe two group homomorphisms $\tau \mapsto \varphi (\tau )$ and $\tau \mapsto {\sigma }^{-1}\tau \sigma$ agree on a generating set of a group, so agree on every product of its elements, hence throughout $S_n$:Finally, $S_1$ is trivial and $S_2$ has a unique nonidentity element, so every automorphism of either is the identity and we can take $\sigma =e$. This completes all degrees $n\ne 6$ and proves that every such automorphism is an inner automorphism.

Assume first $n\ge 2$. All transpositions form a single conjugacy class $T_1$. A group automorphism preserves order of a group element, so the image of a transposition is a nonidentity involution. By the order formula, its disjoint cycle decomposition contains only one-cycles and two-cycles, with at least one two-cycle. Thus it lies in some $T_k$, where $1\le k\le \lfloor n/2\rfloor$.

Moreover an automorphism maps the entire conjugacy class onto the conjugacy class of the image: for a fixed transposition $t$,and $\varphi (g)$ runs over all of $S_n$ by surjectivity. Part (a)(i) identifies that latter conjugacy class as $T_k$. Hence $\boxed{\varphi (T_1)=T_k}$, with equality rather than just inclusion. If $n=1$, $T_1$ is empty and the equality holds trivially with $k=1$.

For distinct $i,j\ne 1$, direct conjugation givesThus the given star transpositions generate every transposition. Every cycle can in turn be expressed asBy the disjoint cycle decomposition, every permutation is a product of cycles, hence of transpositions, hence of the given star transpositions. ThereforeFor $n=1$, the empty generating set of a group generates the trivial group $S_1$.