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Under dependent choice and the ultrafilter lemma, the Samuel compactification of the open unit interval is the closed unit interval

Example

Give $(0,1)$ and $[0,1]$ the subspace metric $d(s,t)=|s-t|$ from the usual real metric (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y) = |x-y|$ is a metric, its open balls are the intervals $(x-r, x+r)$ , and it is unbounded, Isometry, isometric embedding, and the subspace metric on a subset) and the uniformities that it generates. With these metric uniformities, the inclusion $i:(0,1)\to[0,1]$ is a Hausdorff completion. Consequently, under dependent choice and the ultrafilter lemma, $[0,1]$ is the Samuel compactification of $(0,1)$ .

Facts & Assumptions

Given: A real $\varepsilon>0$ and the specified subspace metric uniformities on $(0,1)$ and $[0,1]$ .

[L1]

The metric entourages induce the metric topologies and are separated (A metric on a nonempty set generates an entourage uniformity whose induced topology and uniformly continuous maps are the usual metric notions, and this uniformity is separated).

[L2]

There is $m\ge1$ with $1/m<\varepsilon$ , inverse order reverses on positives, and every real has an integer part (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ , Inverses of positives are positive, and reciprocation reverses order, Integer part: for every real $x$ there is exactly one integer $m$ with $m \le x < m + 1$ ).

[L3]

A set is finite when it is equinumerous with a natural number, and total boundedness asks for a finite entourage-ball cover (The cardinality $\lvert A\rvert$ of a finite set, Totally bounded uniform space).

[L4]

The interval $[0,1]$ is compact, hence complete for its compatible uniformity (Heine-Borel by bisection: every closed bounded interval $[a,b]$ is compact, Every compact uniform space is complete, Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length).

[L5]

Under dependent choice, the Samuel completion of a separated totally bounded space is its ordinary uniform completion; under the ultrafilter lemma it is compact (Under dependent choice the Samuel completion of a separated totally bounded space is its uniform completion; under the ultrafilter lemma it is compact).

Verification

technique · direct

1.1

Choose $n=m+1$ , where $m$ is from [L2]; then $n\ge2$ and $1/n<\varepsilon$ . The map $j\mapsto(j+1)/n$ is injective on the natural number $m$ , so its image $F=\{(j+1)/n:j\in m\}$ is finite and lies in $(0,1)$ .

L2L3

2.1

For $x\in(0,1)$ , write $k=\lfloor nx\rfloor$ . If $k=0$ , then $(0+1)/n\in F$ is within $1/n$ of $x$ . Otherwise $1\le k<n=m+1$ , so $k-1\in m$ , $k/n\in F$ , and $k/n\le x<(k+1)/n$ . Thus $F$ is an $\varepsilon$ -net.

L1L2step 1.1

3.1

Hence $(0,1)$ is separated and totally bounded. The inclusion $i$ pulls back every metric entourage of $[0,1]$ to the same-radius metric entourage of $(0,1)$ , and its image is dense because every interval about $0$ , $1$ , or an interior point meets $(0,1)$ .

L1L3step 2.1

4.1

By [L4], $[0,1]$ is complete, and by [L1] its metric uniformity is separated; so step 3.1 verifies the completion conditions of A Hausdorff completion of a uniform space and its canonical dense map.

L1L4step 3.1

5.1

The identification in [L5] now gives the Samuel compactification $S((0,1))\cong[0,1]$ .

L5step 4.1∎

Depends on

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Sources

Encyclopedia of Mathematics, Uniform space (standard reference, not scraped)