Uniform Spaces: the Three Definitions: Examples and Counterexamples

The entourage $E_{\epsilon }$ is exactly the relation $d(x,y)<\epsilon$.

Applying [L2] gives the usual topology and identifies uniform continuity for these entourages with the usual $\epsilon$-$\delta$ condition.

Direct algebra gives $\mathrm{\mid }h(x)-h(y)\mathrm{\mid }\le 2\mathrm{\mid }x-y\mathrm{\mid }$, so $h$ is uniformly continuous; its displayed inverse and the usual open-interval formulas make it a homeomorphism.

Put $a_n=n\mathrm{/}(n+1)$ and $b_n=(n+1)\mathrm{/}(n+2)$. Then $\mathrm{\mid }{a}_n-b_n\mathrm{\mid }\to 0$, while $\mathrm{\mid }{h}^{-1}(a_n)-h^{-1}(b_n)\mathrm{\mid }=1$.

Thus $h^{-1}$ is not uniformly continuous, so this homeomorphism is not a uniform isomorphism (Uniform embedding and uniform isomorphism, Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological).

Pull the usual uniformity of $(-1,1)$ back along the homeomorphism $h$, obtaining a uniformity $\mathcal{V}$ on the underlying set $\mathbb{R}$.

Since $h$ is a homeomorphism, $\mathcal{V}$ induces the usual topology of $\mathbb{R}$, so that topology is uniformizable.

If $\mathcal{V}=\mathcal{U}$, then $h$ would be a uniform isomorphism from $\mathcal{U}$ onto the usual uniformity of $(-1,1)$, contrary to [L1].

Thus one topology has distinct compatible uniformities, refuting the statement.

In additive notation, both left and right conditions are $y-x\in U$; for symmetric interval neighbourhoods $U=(-\epsilon ,\epsilon )$ this is $\mathrm{\mid }x-y\mathrm{\mid }<\epsilon$.

Hence the left and right uniformities equal the metric uniformity.

Their upper join and Roelcke meet equal that same uniformity because the two inputs already agree.

For fixed $k$, all $n>k$ have $f_n(k)=0$, so the coordinate sequence converges to $0$.

For every $n$, $f_n(n)=1$, so $f_n$ is not in the uniform entourage induced by equality on $\{0,1\}$.

Thus $f_n$ converges pointwise to zero by [L2].

Hence convergence is not uniform.

The interval $[0,1]$ is closed and bounded, hence compact by [L2], and its metric topology is Hausdorff by [L1].

Its restricted metric uniformity is compatible by the metric dictionary (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).

Uniqueness follows from [L3], so no other compatible uniformity exists.

Suppose the $K$-topology were uniformizable.

It would be regular by [L1].

This contradicts its nonregularity in [L2].

Therefore the supposition is false, and the $K$-topology is not uniformizable.

Each two-point factor is compact Hausdorff, so [L1] makes $2^I$ compact and [L7] makes it Hausdorff; [L5] then makes it uniformizable.

At the constant-zero point, suppose $(B_n)$ were a countable local base. For each $n$, choose a finite coordinate set $F_n$ such that the basic zero-cylinder restricting $F_n$ is contained in $B_n$; countable choice licenses these selections. The union ${\bigcup }_n{F}_n$ is at most countable by [L6], so choose $i\in I\setminus {\bigcup }_n{F}_n$ (for instance take $I=\mathcal{P}(\mathbb{N})$, uncountable by [L4]). The one-coordinate neighbourhood requiring coordinate $i$ to be zero contains no $B_n$, because the cylinder inside $B_n$ permits coordinate $i$ to be $1$. This contradicts the local-base property.

Suppose $2^I$ were metrizable. Then [L3] would make it first countable, contradicting step 1.2.

Hence it is not metrizable, and step 1.2 gives failure of first countability.