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4

Hint: if $S$ is a non-empty (bounded) set of real numbers, to show $\sup S \le b$, it is sufficient to show that for any $x \in S$, $x \le b$.


3

Note that the right hand side of the inequality $d(p, q) < 2\epsilon + \operatorname{diam}E$ is a constant independent of $p$ and $q$, so we see that $2\epsilon + \operatorname{diam}E$ is an upper bound for $\{d(p, q) \mid p, q \in \overline{E}\}$. As such, $\operatorname{diam}\overline{E}$, the least upper bound for $\{d(p, q) \mid p, q \in ...


3

We will show that for fixed $q$ the function $\frac{(1+x)^q-1}{x^q+x}$ is bounded. By L'Hospitals Rule, we have $$\lim_{x\to 0^+}\frac{(1+x)^q-1}{x^q+x}=\lim_{x\to 0^+} \frac{q(1+x)^{q-1}}{qx^{q-1}+1}=q.$$ And it is clear that $$\lim_{x\to \infty}\frac{(1+x)^q-1}{x^q+x}=1$$ if $q\gt 1$. (The case $q=1$ is easy to deal with.) Boundedness follows.


3

The metric on the subspace is just the restriction of the metric from the original space. So $d(2,5)$ in the space you've written is $3$. There is not much you can say that is special about this situation from the metric point of view. One thing you can say is that if $A,B$ are as above and $d(a,b)$ is bounded below for $a \in A$ and $b \in B$, then $X$ is ...


2

If you notice that each of these points is isolated, you could observe that this set is the complement of the set $$\bigcup_{n=1}^\infty (-(n+1)+\frac{1}{(n+1)},-n+\frac{1}{n}) \cup (0,\infty)$$ Which is an infinite union of open sets and therefore open (and whose complement is closed).


2

A set is closed if and only if it contains all of its limit points. Your set has no limit points, so it's trivially closed by that criteria. To explicitly show that your set has no limit points, note that for any $x \in \mathbb{R}$, the open set $(x-1/2,x+1/2)$ contains at most one point in $M$ since distance between successive points of $M$ is $$\left(-n + ...


2

So we are asked to find: $$ \sum_{j=0}^{+\infty}\frac{1}{j+1}\binom{2j}{j}\int_{0}^{1/n}x^{j+1}(1-x)^{j}\,dx.\tag{1}$$ Since: $$ \sum_{j\geq 0}\frac{1}{j+1}\binom{2j}{j}z^j = \frac{1-\sqrt{1-4z}}{2z}\tag{2}$$ we just need to find: $$ \int_{0}^{1/n} x\cdot\frac{1-\sqrt{1-4x(1-x)}}{2x(1-x)}\,dx = \int_{0}^{1/n}\frac{1-|1-2x|}{2(1-x)}\,dx\tag{3}$$ and assuming ...


1

In fact you will be able to prove that if $\lim\limits_{n \to +\infty} -k_n + \frac{1}{k_n}$ converges to a real $x_0$ then $(k_n)$ has to be eventually constant. Hence $x_0 \in M$ as was supposed to be proven. Another way to prove the result is to prove that the complement of $M$ is open as mentionned in the other response.


1

Since $\nabla g$ is non-vanishing in $A$ by hypothesis, the normalized gradient $\nu = \nabla g/\|\nabla g\|$ is a continuous unit normal field on the hypersurface $M$. Since the ambient space is orientable, $M$ itself is orientable.


1

Suppose $y(0) = y_0; \tag{1}$ then the unique solution to the equation $y' = a(x)y + b(x) \tag{2}$ is $y(x) = e^{\int_0^x a(s)ds}(y_0 + \int_0^x e^{-\int_0^s a(u) du} b(s) ds); \tag{3}$ formula (3) is very well known; a derivation may be found here; for $x = T$ we thus have $y(T) = e^{\int_0^T a(s) ds}(y_0 + \int_0^T e^{-\int_0^s a(u) du} b(s) ds). ...


1

http://mathworld.wolfram.com/IntegratingFactor.html I am suppressing $(x)$ in $a(x)$ etc. $$y=e^{\int a\, dx}\big(\int b\,e^{-\int a\, dx}dx +C \big)$$ Let $\int_0^T a\, dx = M$ and $\int_0^T b\, dx = N$. Then we can write $\int a\, dx=A(x)+\frac MTx$ and $\int b\, dx=B(x)+\frac NTx$ where $A$ and $B$ are zero mean AND periodic. FACT: Integral of a zero ...


1

Yes. There could be for some manifold. For example if you have a Killing vector fields $X$ on $M$, then the diffeomorphism group generated by $X$ are isometries. Thus it sends geodesics to geodesics. It is known that such a vector fields are Jacobi fields (along any geodesics). For example, any constant vector fields on $\mathbb R^n$ and flat torus are ...



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