I.M.James' proof of "uniformizable spaces are completely regular"

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In I.M. James' Topological and uniform spaces, the proof for the second half of Proposition 11.5, on p.143, makes me confused. There, he proves a unformizable space is completely regular.

The outline of his proof is, for a uniform space $X$, a closed subset $H$ and a point $x\in X\setminus H$, he wants to find a continuous function $\alpha: X\to [0,1]$ such that $\alpha=0$ on $H$ and $\alpha(x)=1$. Since $X$ is uniform, consider a sequence of entourages $\{D_i: i=0, \cdots\}$ such that $D_0[x]\subset X\setminus H$ and $D_{i+1}\circ D_{i+1}\subset D_i$, for $i=0,\cdots$. Then, for every $t\in I'$, where $I'$ is the set of dyadic rationals in $[0,1]$, he defines $E_t:= D_{i_n}\circ \cdots \circ D_{i_1}$, where $i_k$'s are the nozero digits of binary form of $t$. That is, $$t=\frac{1}{2^{i_1}}+\cdots + \frac{1}{2^{i_n}}.$$ Then he defines the function $\alpha$ as
$$\alpha(\xi):=\begin{cases} \inf\{t\in I': \xi \notin E_{t}[x]\},& \text{for }\xi\neq x\\ 1,& \text{for } \xi =x\end{cases}$$ and finally shows it is uniformly continuous.

But function $\alpha$ seems contradictory. Because the size of $E_t[x]$ increases as dyadic $t$ does.

Am I misunderstanding something? it seems to me the correct definition should be $$\alpha(\xi):=\begin{cases} \inf\{t\in I': \xi \notin E_{1-t}[x]\},& \text{for }\xi\neq x\\ 1,& \text{for } \xi =x\end{cases}.$$