Let $\mathcal G$ be a bounded-degree (Borel) pmp graph on a standard probability space $(X, μ)$. The set $L^2(X)$ consists of all functions $f : X → \mathbb C$ such that
$$ \int_{x\in X} |f(x)|^2\ d\mu(x) <\infty $$
Define the adjacency operator $T_\mathcal G$ of $\mathcal G$ as follows. For each $f ∈ L^2(X)$ and for each $x ∈ X$,
$$ T_{\mathcal G}(f)(x) = \sum _{y\sim x} f(y) $$
where $y ∼ x$ means “$y$ is adjacent to $x$”. Prove that, if $\mathcal G$ is a finite graph, then the adjacency operator coincides with the usual adjacency matrix. Hint: Any vector in $\mathbb C^n$ can be identified with a function $\{1, . . . , n\} → \mathbb C$.
If $\mathcal G$ is finite, $X$ is finite, so we can enumerate and relabel its vertices $X=\{1,\dots,n\}$ while preserving the probability measure. Then $L^2(X)$ contains all functions $f : X → \mathbb C$, i.e. $L^2(X)=\mathbb C^X$
$$ \forall f\in \mathbb C^X\quad \int_{x\in X} |f(x)|^2\ d\mu(x) =\sum_{x\in X} |f(x)|^2\cdot \mu(x) <\infty
$$
Moreover, $\mathbb C^X$ can be identified with $\mathbb C^n$ by simply organizing the outputs of each function $f\in \mathbb C^X$ as a vector $\vec v\in \mathbb C^n$ through the linear bijection
$$ h:\mathbb C^X\to \mathbb C^n: f\to \vec v_f,\quad \vec v_f(i)=f(i) $$
Consider the adjacency matrix $A$ of $\mathcal G$ with respect to this enumeration of $X$. The $i$th row of $A$ is given by $\vec w^t$ where
$$ \vec w(j)=\begin{cases} 1&\text{if $i\sim j$}\\ 0&\text{otherwise} \end{cases} $$
So for any vector $\vec v\in \mathbb C^n$, $A\vec v$ can be characterized by:
$$ \forall i\in\{1,\dots,n\}\quad A\vec v(i)= \vec w^t \vec v= \sum_{j=1}^n \vec w(i)\cdot \vec v(j)= \sum_{j \sim i}\vec v(j) $$
which coincides with the definition of the adjacency operator when we identify each vector $\vec v \in \mathbb C^n$ with a function $f\in \mathbb C^X$.
Let $\mathcal G$ be a bounded-degree (Borel) pmp graph on a standard probability space $(X, μ)$. The Laplacian operator $L_{\mathcal G}$ of $\mathcal G$ is defined as follows. For each $f ∈ L^2(X)$ and for each $x ∈ X$,
$$ L_{\mathcal G}(f)(x) = \deg(x)f(x)-\sum _{y\sim x} f(y) $$
Prove that $L_{\mathcal G}$ is a bounded operator.
We can express $L_{\mathcal G}$ as the difference of two bounded operators:
$$ L_\mathcal G = D-T_\mathcal G,\quad \text{where } (Df)(x) = \deg(x)\cdot f(x) $$
We have already seen that $T_\mathcal G$ is bounded. $D$ is also bounded as an operator, since if $d$ is the maximal degree of $\mathcal G$:
$$ |(Df)(x)|=|\deg(x)||f(x)| \le d\cdot |f(x)|\\[4pt] \Rightarrow \quad
\int \left| \left(Df\right)(x) \right|^2 \ dx \le
\int d^2\cdot \left| f(x) \right|^2\ dx
=d^2{\|f\|_2}^2 \\[4pt] \Rightarrow \quad
\forall f\in L^2(X)\quad \frac{\|D f\|_2} {\|f\|_2} \le d
\quad\Rightarrow \quad \left\|D\right\|\le d $$
So by triangle inequality,
$$ \left\|L_\mathcal G\right\| \le \left\|T_\mathcal G\right\| +\left\|D\right\| \le 2d $$
Let $\R$ be the space of real numbers equipped with the usual topology. Prove that $[0,1)$ is neither open nor closed but is both $G_δ$ and $F_σ$.
The usual topology on $\R$ is compatible with the standard metric $d(x,y)=|x-y|$. So $[0,1)$ is not open since $0\in[0,1)$ but there is no open ball centered at $0$ contained in $[0,1)$. And $[0,1)$ is not closed since its complement $(-\infty,0)\cup[ 1,\infty)$ is not open for a similar reason—there is no open ball centered at $1$ contained in $(-\infty,0)\cup[ 1,\infty)$.
But
$$ [0,1)= \bigcap _{n=1}^\infty
\left(-\frac 1 n, 1\right) ,\\[4pt] [0,1)= \bigcup _{n=1}^\infty
\left[0, 1-\frac 1 n\right]
$$
So indeed $[0,1)$ is $G_δ$ and $F_σ$.
Let $\R$ be the space of real numbers equipped with the usual topology, and view $(0, 1)$ as a subspace of $\R$. Prove that $(0, 1)$ is a Polish space.
$(0,1)$ is separable as has the countable dense subset $(0,1)\cap \mathbb Q$. While $(0,1)$ is not complete with respect to the usual metric on $\R$, it is complete with respect to the metric
$$ d(a,b)=\left| g(a)-g(b) \right|,\quad g(x)=\tan\left(\pi x+\frac{\pi}{2}\right) $$
Visually, this metric is the result of:
representing $a$ and $b$ on the lower semicircle
$$ H = \left\{ \left(x,-\sqrt{1-\ x^{2}}+1\right) :x\in (0,2\pi) \right\} $$
through a linear parameterization:
$$ f:(0,1)\to H: x\mapsto \left(\cos(\pi x+\pi), \sin(\pi x+\pi)+1\right) $$
projecting through the point $(0,1)$ and $a$ and $b$ onto the real line
taking the distance between the projected points
https://www.desmos.com/calculator/vfvrq6fj1n
The topology under this metric is generated by the set of balls:
$$ \mathcal B=\{B_r(x)\mid x\in(0,1),\ r\in \R\} $$
Each of these balls is simply an open interval on $(0,1)$, for example, $B_2(0.3)\approx (0.11,0.79)$:
https://www.desmos.com/calculator/47osijcsyl
Conversely, any open interval on $(0,1)$ can be represented as a ball with respect to this new metric. (By projecting the interval onto $\R$ using the steps outlined previously, you can find an appropriate center and radius.)
Thus, the set of open intervals is exactly equal to $\mathcal B$; i.e. this metric is compatible with the subspace topology.
So $(0,1)$ with the subspace topology is completely metrizable and seperable, i.e. it is Polish.
Let $X$ be a Polish space. Prove that $X$ has a countable basis.
If $X$ is Polish, it is separable and completely metrizable with respect to some metric $d:X\times X\to \R$, i.e. $(X,d)$ is a metric space.
$X$ has a countable dense subset $S$. Consider the countable collection of open balls:
$$ \mathcal B = \{B_r(s)\mid s\in S, r\in \mathbb Q_+\} $$
We will prove $\mathcal B$ is a base for the topology of $X$ (and therefore $X$ is second-countable).
Let $U\in \mathcal T$ be a nonempty open set and $x\in U$. It suffices to show $\exist B\in \mathcal B$ such that $x\in B\sube U$.
Since $U$ is open, there is some sufficiently small rational $\epsilon$ such that $B_\epsilon(x)\sube U$. And as $S$ is dense in $X$, there exists $s\in S$ such that $d(s,x)<\frac \epsilon 2$.
Then, by triangle inequality $\forall y\in B_\frac \epsilon2 (s)$
$$ d(y,x)\le d(y,s)+d(s,x)<\frac \epsilon 2+\frac \epsilon 2 =\epsilon \quad \Rightarrow \quad y\in B_\epsilon(x) $$
So
$$ B_{\frac\epsilon 2}(s)\sube B_\epsilon(x)\sube U $$
And as $\frac \epsilon 2$ is rational, $B_{\frac\epsilon 2}(s)\in \mathcal B$ with
$$ x\in B_{\frac\epsilon 2}(s) \sube U $$
as desired.
Let $f : \R → \R$ be a function, and let $C = \{x ∈ \R : f\text{ is continuous at }x\}$. Prove that $C$ is a Borel set.
We can express $C$ as the countable intersection
\inf_{r>0} \sup_{y,z\in B_r(x)} |f(y)-f(z)| $$
Notice that for any $k$, any $x\in C_k$,
$$
\inf_{r>0} \sup_{y,z\in B_r(x)} |f(y)-f(z)| <\frac 1k $$
Since this is the case, there most be some sufficiently small $\epsilon$ such that
$$ \sup_{y,z\in B_\epsilon(x)} |f(y)-f(z)|<\frac 1k $$
and since $B_\epsilon(x)$ is open, for any point $x'\in B_\epsilon(x)$, there is some sufficiently small $\delta$ such that $B_\delta(x')\sube B_\epsilon(x)$ and therefore
\inf_{r>0} \sup_{y,z\in B_r(x')} |f(y)-f(z)| <\frac 1k $$
i.e. $x'\in C_k$ and therefore $B_\epsilon(x)\sube C_k$. So $C_k$ is open for all $k$, and therefore $C$ is $G_\delta$. This forces $C$ to be Borel.