Property of join

 Let $X$ be path-connected and $Y$ be arbitrary topological space. Then the join $X*Y$ is simply connected.

 

$\textbf{Proof}.$ We use Van Kampen style argument. Define $U=X*Y\setminus Y=X\times Y\times [0,1)/\sim,$ where $\sim: \ (x,y_1,0)\sim(x,y_2,0) $ for all $y_1,y_2\in Y$.  Thus $U$ is clearly open and deformation retracts onto $X\cong  X\times Y\times \{0\}/\sim.$ As $X$ is connected, so is $U$.

Now fix $x_0\in X$. Let $V_1$ be the cone with apex $x_0$ and base  $\{x_0\}\times Y\times\{1/2\}$. Also define $V_2:=X\times Y\times [1/2,1]/\sim$, where $\sim: \ (x_1,y,0)\sim(x_2,y,0) $ for all $x_1,x_2\in X.$ Set $V=V_1\cup V_2$. Observe that $V_2$ deformation retracts onto  $\{x_0\}\times Y\times\{1/2\}$ (collapse everything to $Y$ and then push it to $\{x_0\}\times Y\times\{1/2\}$). But this is the base of the cone $V_1$, and $V_1$ deformation retracts to $x_0$. Hence $V$ is contractible.

The set $V$ is not open, but $U\cup \text{int}V=X$. Moreover, $U\cap V=V_1\cup (X\times Y\times [1/2,1))$ is connected. Thus every loop in $X*Y$ based at $x_0$ is homotopy equivalent to finite product of loops at $x_0$, each belonging to $U$ or $V$. Observe that $X$ is nullhomotopic in $X*Y$ (use a cone with base $X$ and apex some point in $Y$), and hence $U$ is such. Thus the inclusion $U\hookrightarrow X*Y$ induces a trivial homomorphism $\pi_1(U)\to \pi_1(X)$. Moreover, $\pi_1(V)$ is already trivial. Thus each loop in $X*Y$ is homotopy equivalent to a constant loop.

Homotopy type of R^3\(circles)

$\mathbf{1}$. The spaces $X=\mathbb{R}^3\setminus S^1$ and $S^{1}\vee S^{2}$ are homotopy equivalent.

First we rewrite $X\cong S^3\setminus (\{*\}\cup S^1)$, as $S^3$ is the one-point compactification of $\mathbb{R}^3$. Now observe $S^3\setminus S^1$ is homeomorphic to the open solid torus, $S^1\times e^2$, where $e^2$ is the open unit $2$-disc. An explicit homemorphism is as follows : $$(x,y,z,t)\mapsto (x\sqrt{1-z^2-t^2},y\sqrt{1-z^2-t^2},z,t).$$

 Thus $\mathbb{R}^3\setminus S^1\cong S^1\times e^2\setminus \{*\}$. Here $S^1\times e^2$ is open $3$-manifold. Removing a point from open $3$-manifold $M$ is wedging $M$ with $S^2$ $-$ we may think of enclosing the point with $S^2$ and retracting the enclosed region to the sphere (or blowing up the point to a sphere); then we may push the sphere away from the manifold, to the boundary, forming the wedge product. Thus $\mathbb{R}^3\setminus S^1\cong S^1\times e^2\vee S^2\sim S^1\vee S^2$. 

 

$\mathbf{2}$. The space $X=\mathbb{R}^3\setminus M$, where $M$ is the union of two disjoint unlinked circles, is homotopy equivalent to $S^{1}\vee S^1\vee S^{2}\vee S^2$.

As before, $X\cong S^1\times e^2\setminus(\{*\}\cup S^1)$. Observe that the remaining circle to be removed is contractible in $S^1\times e^2$, i.e. it is not encircling the hole of the torus. Thus we may take a $2$-sphere which bounds a region containing the circle to be removed. This regions retracts to $S^2\vee S^1$, as in $\mathbf{1}$. Pushing it to the boundary of $S^1\times e^2$ we get a wedge with $S^2\vee S^1.$ Removing the point $*$ induces, as in $\mathbf{1}$, one more wedge, with $S^2$. Hence $X\sim (S^1\times e^2)\vee S^2\vee S^2\vee S^1$, which finishes the derivations as $S^1\times e^2\sim S^1$

 

$\mathbf{3}$. The space $X=\mathbb{R}^3\setminus M$, where $M$ is the union of two disjoint linked circles, is homotopy equivalent to $S^{2}\vee (S^1\times S^1)$.

As in $\mathbf{2}$, $X\cong S^1\times e^2\setminus(\{*\}\cup S^1)$. Observe that the remaining circle to be removed is not contractible in $S^1\times e^2$, i.e. it encircles the hole of the torus. Removing the point $*$ induces, as in $\mathbf{1}$, a wedge with $S^2$.  Hence $X\sim ((S^1\times e^2)\setminus S^1)\vee S^2$. But $(S^1\times e^2)\setminus S^1$ is a solid torus with core circle removed, which clearly deformation retracts to a torus $-$ $S^1\times S^1$.

 

 

 

No retraction

 Let $I=[0,1]$, $A=\{0\}\cup\{\frac{1}{n}\}_{n\ge 1}$. Prove that there is no retraction from $I\times I$ to $I\times \{0\}\cup A\times I$. 

OMOUS 2024

Let $A$ be a real matrix such that $A+A^2A^t+(A^t)^2=0$. Prove that $A=0$.

$\textbf{Solution}.$ Multiply the given equation by $A^t$ from the right to get $AA^t+A^2(A^2)^t+(A^t)^3=0$ and hence $(A^t)^3=-AA^t-A^2(A^2)^t$ . Observe that RHS is negative semi-definite symmetric matrix. Thus $(A^t)^3$ and hence $A^3$ is symmetric negative semi-definite.

Now multiply the given equation with $A$ from the left to get  $A^2+A^3A^t+A(A^t)^2=0$ and using $A^3=(A^t)^3$ we get $A^2+(A^t)^4+A(A^t)^2=0$. Transposing this equation we get $(A^t)^2+A^4+A^2 A^t=0$. Comparing with the original equation we get $A^4=A$. Thus $A$ is diagonalizable and $\lambda^4=\lambda$ for every characteristic root $\lambda$ of $A$. Thus  $\lambda=0$ or $\lambda^3=1$. But $A^3$ is negative semi-definite, so $\lambda^3\le 0$. Thus $\lambda=0$ and hence all roots of $A$ are $0$. Since $A$ is diagonalizable, $A=0$.

Truncated exponential series equation

$\textbf{Problem}.$ For a positive integer $n$ let $x_n$ be the unique positive real root of the equation

$$\sum_{k=0}^n \frac{x^k}{k!}=\frac{e^x}{2}.$$

Prove that $\displaystyle\lim_{n\to\infty}(x_n-n)=\frac{2}{3}.$

$\textbf{Comment.}$ The limit $\displaystyle\lim_{n\to\infty} e^{-n}\sum_{k=0}^n \frac{n^k}{k!}$ is well known to be $\displaystyle\frac{1}{2}$. It could be derived via nice probabilistic structure related to Poisson distribution. This limit hints that the value of the root $x_n$ should be close to $n$. 

$\textbf{Sketch of a solution of the problem}$. Defining 

$$g(x)=\frac{1}{n!}\int_0^x s^ne^{-s}\ ds-\frac{1}{2}$$ 

the original equation could be rewritten in the form $g(x)=0.$ One should observe that $g$ is monotonically increasing and the unique root is in the interval $(n,n+1)$. Then one uses one iteration of the Newton method initialized at $n$ to obtain an approximation of the solution $\tilde x_n$. Using the final asymptotic obtained at the end of the this answer and Stirling approximation, one obtains that $\displaystyle\lim_{n\to\infty}(\tilde x_n-n)=\frac{2}{3}.$ It remains to use the estimate of the error in the Newton method (one needs the first and second derivatives of $g$ here, which are easy to write), to see that $\lim_{n\to\infty}(x_n-\tilde x_n)=0$.

Proof of Gelfand-Phillips Theorem

$\textbf{Theorem.}$ Let $X$ be a Banach space and $A\subseteq X$. Prove that $A$ is precompact (in the norm) if and only if for every $w^*$-convergent to $\textbf{0}$ sequence $\{x_n^*\}_{n\ge 1}\subseteq X^*$ it holds that $\{x_n^*\}_{n\ge 1}$ converges uniformly to $\textbf{0}$ on  $A.$

 

$\textbf{Proof.}$ If $A$ is precompact the result follows from Arzela-Ascoli theorem (combined with the fact that $w^*$-convergent sequences are norm bounded).

Now we prove the reverse direction. 

First we proof the following characterization of compactness in normed spaces: $A$ is compact if and only if $A$ is bounded and for any $\varepsilon>0$ there exists a finite-dimensional space $F$ such that $A\subseteq F+\varepsilon\mathbf{B}.$

$\textit{Proof}.$ 

If $A$ is precompact, then for any $\varepsilon>0$ there exists $\{a_i\}_{i=1}^n\subseteq A$ such that $A\subseteq\bigcup_{i=1}^n\mathbf{B}_{\varepsilon}(a_i)$. In particular $A$ is bounded and
\[A\subseteq \text{span}(\{a_i\}_{i=1}^n)+\varepsilon\textbf{B}.\]
For  the reverse direction, let $\varepsilon>0$ and $F$ be finite-dimensional space such that $A\subseteq F+\frac{\varepsilon}{3}\mathbf{B}.$ For each $a\in A$ choose $f_a\in F$ with $\|a-f_a\|\le\varepsilon/3$. Since $A$ is bounded and $F$ is finite-dimensional, the set $\{f_a\}_{ a\in A}$ is precompact.
Thus there exists $\{f_{a_i}\}_{i=1}^n\subseteq F$ such that $ \{f_a\}_{a\in A}\subseteq \bigcup_{i=1}^n\textbf{B}_{\varepsilon/3}(f_{a_i})$.
Now let $a\in A$ be arbitrary. Then there exists $i\in \{1,2,\ldots,n\}$ such that $\|f_a-f_{a_i}\|<\varepsilon/3.$ Consequently
\[\|a-a_i\|\le\|a-f_a\|+\|f_a-f_{a_i}\|+\|f_{a_i}-a_i\|<\varepsilon,\]
hence $\{a_i\}_{i=1}^n\subseteq A$ is a finite $\varepsilon$-net for $A.$ $\square$

Now let $A$ be a set which is not precompact. If it is unbounded, the results follows easily. Assume otherwise. Thus there exists $\varepsilon>0$ such that for any finite-dimensional space $F$, there exists $a\in A$ with $\text{d}(a,F)\ge\varepsilon.$ Choose a sequence $\{x_i\}_{i\ge 1}\subseteq X$ which is dense in $X$. Let $F_n:=\text{span}(\{x_i\}_{i=1}^n).$ Choose $a_n\in A$ with $\text{d}(a_n,F_n)\ge\varepsilon.$ Using Hahn-Banach construct $x_n^*\in X^*$ such that $\langle x_n^*,a_n\rangle\ge\varepsilon$, $\|x_n^*\|=1$ and $F_n\subseteq\text{ker}(x_n^*)$.
Thus for every $n$ \[\sup_{a\in A}|\langle x_n^*,a\rangle|\ge\varepsilon,\]
hence $\{x_n^*\}_{n\ge 1}$ does not converge uniformly to $\mathbf{0}$ on $A$. On the other hand $\bigcup_{n\ge 1}F_n$ is dense in $X$. Fix $x\in X$ and let $\delta>0$ be arbitrary. Then there exists $m$ such that for some $z_m\in F_m$ it holds that $\|z_m-x\|<\delta$. For $n\ge m$ it holds that $\langle x_n^*,z_m\rangle=0$, hence
\[|\langle x_n^*,x\rangle|= |\langle x_n^*,z_m-x\rangle|\le \|x_n^*\|\|z_m-x\|<\delta.\]
Consequently $\{x_n^*\}_{n\ge 1}$ is weak$^*$ convergent to $\mathbf{0}.$

Closed orbits are finite

 Let $f:[a,b]\to [a,b]$ be continuous function and $x_0\in [a,b]$. Consider the sequence $X=\{x_n\}_{n\ge 0}$ defined by $x_n=f(x_{n-1})$ for $n\ge 1$. Assume that $X$ is a closed set. Prove that it consists of finitely many elements. (The problem is taken from IMC 2002. The set $X$ is called the orbit of $f$ starting from $x_0$)

$\textbf{Proof.}$ Let us assume that $X$ is infinite. So it has a limit point, which should belong to the set (since it is closed). Let $$n_0:=\min\{n\ | \ x_n \text{ is a limit point of } X \}.$$ Then define $\hat X=\{x_n\}_{n\ge n_0+1}$. Thus $\hat X\cup \{x_{n_0}\}$ is compact and $f:\hat X\cup\{x_{n_0}\}\to \hat X$ is a continuous function. Since the image of compact sets under continuous functions are compact set, we must have that $\hat X$ is compact as well. But it is clearly not closed, since $x_{n_0}$ is a limit point $\hat X$ and does not belong to $\hat X$.