In that article, I described some properties of Thomae’s functionf. Namely:
The function is discontinuous on Q.
Continuous on R∖Q.
Its right-sided and left-sided limits vanish at all points.
Let’s modify f to get function g defined as follow: g:|R⟶Rx⟼0 if x∈R∖Qpq⟼q if pq in lowest terms and q>0f and g both vanish on the set of irrational numbers, while on the set of rational numbers, g is equal to the reciprocal of f. We now consider an open subset O⊂R and x∈O. As f right-sided and left-sided limits vanish at all points, we have limn→+∞f(xn)=0 for all sequence (xn) of rational numbers converging to x (and such a sequence exists as the rational numbers are everywhere dense in the reals). Hence limn→+∞g(xn)=+∞ as f is positive.
We can conclude that g is nowhere locally bounded. The picture of the article is a plot of function g on the rational numbers r=pq in lowest terms for 0<r<1 and q≤50.
Can the symmetric groupSn be generated by any transposition and any n-cycle for n≥2 integer? is the question we deal with.
We first recall some terminology:
Symmetric group
The symmetric group Sn on a finite set of n symbols is the group whose elements are all the permutations of the n symbols. We’ll denote by {1,…,n} those n symbols.
Cycle
A cycle of length k (with k≥2) is a cyclic permutationσ for which there exists an element i∈{1,…,n} such that i,σ(i),σ2(i),…,σk(i)=i are the only elements moved by σ. We’ll denote the cycle σ by (s0s1…sk−1) where s0=i,s1=σ(i),…,sk−1=σk−1(i).
Transposition
A transposition is a cycle of length 2. We denote below the transposition of elements a≠b by (ab) or τa,b.
Let’s investigate the following question: “Is the distance between a point and a hyperplane always reached?”
In order to provide answers to the question, we consider a normed vector space(E,‖ and a hyperplane H of E. H is the kernel of a non-zero linear form. Namely, H=\{x \in E \text{ | } u(x)=0\}.
The case of finite dimensional vector spaces
When E is of finite dimension, the distance d(a,H)=\inf\{\Vert h-a \Vert \text{ | } h \in H\} between any point a \in E and a hyperplane H is reached at a point b \in H. The proof is rather simple. Consider a point c \in H. The set S = \{h \in H \text{ | } \Vert a- h \Vert \le \Vert a-c \Vert \} is bounded as for h \in S we have \Vert h \Vert \le \Vert a-c \Vert + \Vert a \Vert. S is equal to D \cap H where D is the inverse image of the closed real segment [0,\Vert a-c \Vert] by the continuous map f: x \mapsto \Vert a- x \Vert. Therefore D is closed. H is also closed as any linear subspace of a finite dimensional vector space. S being the intersection of two closed subsets of E is also closed. Hence S is compact and the restriction of f to S reaches its infimum at some point b \in S \subset H where d(a,H)=\Vert a-b \Vert. Continue reading Distance between a point and a hyperplane not reached→
From the mean value theorem, a real function whose derivative is strictly positive at every point of an interval is strictly increasing. In particular, a continuously differentiable function f defined in a non-degenerate interval I with a strictly positive derivative at a point a of the interval is strictly increasing near that point. For the proof, we just have to notice that as f^\prime is continuous and f^\prime(a) > 0, f^\prime is strictly positive within an interval J \subset I containing a. By the mean value theorem, f is strictly increasing on J.
We now suppose that f is differentiable on an interval I containing 0 with f^\prime(0)>0. For x>0 sufficiently close to zero we have \displaystyle \frac{f(x)-f(0)}{x-0} > \frac{f^\prime(0)}{2}>0, hence f(x)>f(0). But that doesn’t imply that f is strictly increasing in a neighborhood of zero. Let’s prove it with a counterexample. Continue reading A function whose derivative at 0 is one but which is not increasing near 0→
In this article, we consider a groupG and two subgroups H and K. Let HK=\{hk \text{ | } h \in H, k \in K\}.
HK is a subgroup of G if and only if HK=KH. For the proof we first notice that if HK is a subgroup of G then it’s closed under inverses so HK = (HK)^{-1} = K^{-1}H^{-1} = KH. Conversely if HK = KH then take hk, h^\prime k^\prime \in HK. Then (hk)(h^\prime k^\prime)^{-1} = hk(k^\prime)^{-1}(h^\prime)^{-1}. Since HK = KH we can rewrite k(k^\prime)^{-1}(h^\prime)^{-1} as h^{\prime \prime}k^{\prime \prime} for some new h^{\prime \prime} \in H, k^{\prime \prime} \in K. So (hk)(h^\prime k^\prime)^{-1}=hh^{\prime \prime}k^{\prime \prime} which is in HK. This verifies that HK is a subgroup. Continue reading Two subgroups whose product is not a subgroup→
In that article, I gave examples of real valued functions defined on (0,+\infty) that converge to zero and whose derivatives diverge. But those functions were not monotonic. Here I give an example of a decreasing real valued function g converging to zero at +\infty and whose derivative is unbounded.
We consider a vector space V of dimension 2 over a field \mathbb{K}. The matrix: A=\left( \begin{array}{cc}
0 & 1 \\
0 & 0 \end{array} \right) has several wonderful properties!
Only zero as eigenvalue, but minimal polynomial of degree 2
