precompactness and boundedness in uniform space

Question

Consider a uniform space $(X,\mathscr{U})$. For an entourage $U\in\mathscr{U}$, one says that a set $M$ is small of order $U$, if $M\times M\subseteq U$.

$P\subset X$ is precompact if for every $U\in\mathscr{U}$, $P$ can be covered by finitely many sets which are small of order $U$.

I have stumbled upon a small lemma which I don't know why it's true. It says the following:

a) If $A$ is precompact subset of $(X,\mathscr{U})$, then $A$ is bounded in $X$.

b) If $(X,\mathscr{U})$ is uniformly locally precompact, then $A$ is bounded in $X$ iff $A$ is precompact

c) If $(X,\mathscr{U})$ is uniformly locally compact, then $A$ is bounded in $X$ iff $A$ is relatively compact

If someone can help me understand why these statements are true I will appreciaete it. Thank you in advance for all the help.

Answer 1

For (a), suppose that $A$ is precompact, and let $U\in\mathscr{U}$. For some $n\in\Bbb Z^+$ there are sets $M_1,\ldots,M_n$ in $X$ such that $A\subseteq\bigcup_{k=1}^nM_k$, and $M_k\times M_k\subseteq U$ for $k=1,\ldots,n$.

For $k=1,\ldots,n$ fix $x_k\in M_k$; then $U[x_k]\supseteq M_k$, so if $F=\{x_1,\ldots,x_n\}$, we have

$$U[F]\supseteq\bigcup_{k=1}^nM_k\supseteq A\;,$$

and it follows that $A$ is bounded.

For (c) suppose that $X$ is uniformly locally compact, and fix $U\in\mathscr{U}$ such that $U[x]$ is compact for each $x\in X$. There is a $V\in\mathscr{U}$ such that $V\circ V\subseteq U$, and since the symmetric, closed members of $\mathscr{U}$ are a base for $\mathscr{U}$, we may further assume that $V$ is closed. Thus, $(V\circ V)[x]$ is compact for each $x\in X$.

Proposition. If $K\subseteq X$ is compact, then $V[K]$ is compact.

Proof. Let $K\subseteq X$ be compact. The interiors of the sets $V[x]$ for $x\in K$ cover $K$, so there is a finite $F\subseteq K$ such that $K\subseteq\bigcup_{x\in F}V[x]$. It follows that $V[K]\subseteq\bigcup_{x\in F}(V\circ V)[x]$, where $\bigcup_{x\in F}(V\circ V)[x]$, being a finite union of compact sets, is compact. Suppose that $y\in X\setminus V[K]$. Then $(X\times X)\setminus V$ is a nbhd of $K\times\{y\}$, and the sets $\{y\}$ and $K$ are compact, so there are nbhds $W_0$ of $K$ and $W_1$ of $y$ such that $W_0\times W_1\subseteq(X\times X)\setminus V$. Clearly $W_1\cap V[K]=\varnothing$, so $y\notin\operatorname{cl}V[K]$. Thus, $V[K]$ is a closed subset of the compact set $\bigcup_{x\in F}(V\circ V)[x]$ and is therefore itself compact. $\dashv$

Let $A\subseteq X$ be bounded. There are a finite $F\subseteq A$ and $n\in\Bbb Z^+$ such that $V^n[F]\supseteq A$. $F$ is compact, so it follows by induction from the proposition that $V^k[F]$ is compact for all $k\in\Bbb Z^+$ and hence that $V^n[F]$ is compact. Thus, $\operatorname{cl}A\subseteq\operatorname{cl}V^n[F]=V^n[F]$, and $\operatorname{cl}A$ is therefore compact, as desired.

Now let $A\subseteq X$ be compact, and let $W\in\mathscr{U}$. The interiors of the sets $W[x]$ for $x\in A$ cover $A$, so there is a finite $F\subseteq A$ such that $A\subseteq\bigcup_{x\in F}W[x]$, and it follows that $A$ is bounded.

One direction of (b) is covered by (a). For the other direction let $U\in\mathscr{U}$ be such that $U[x]$ is precompact for each $x\in X$. There is a $V\in\mathscr{U}$ such that $V\circ V\subseteq U$, so we may assume that $(U\circ U)[x]$ is precompact for each $x\in X$.

Proposition. If $K\subseteq X$ is precompact, then so is $U[K]$.

Proof. Let $V\in\mathscr{U}$ be arbitrary, and let $W=U\cap V$. There are sets $M_1,\ldots,M_n$ in $X$ such that $K\subseteq\bigcup_{k=1}^nM_k$, and $M_k\times M_k\subseteq W$ for $k=1,\ldots,n$. The sets $M_k$ are precompact, since $W\subseteq U$. For $k=1,\ldots,n$ fix $x_k\in M_k$, and note that $U[x_k]\supseteq W[x_k]\supseteq M_k$. Each $(U\circ U)[x_k]$ is precompact, so there are $M_{k,1},\ldots,M_{k,m_k}\subseteq X$ such that $(U\circ U)[x_k]\subseteq\bigcup_{j=1}^{m_k}M_{k,j}$, and each $M_{k,j}\times M_{k,m}\subseteq W$. But then

$$U[K]\subseteq\bigcup_{k=1}^nU[M_k]\subseteq\bigcup_{k=1}^n(U\circ U)[x_k]\subseteq\bigcup_{k=1}^n\bigcup_{j=1}^{m_k}M_{k,j}\;,$$

where each $M_{k,j}$ is $\mathscr{W}$-small and hence $\mathscr{V}$-small. $\dashv$