# Definition of uniform subspace

I would like some help with the definition of a uniform subspace.

The textbook I refer to is Topological and Uniform Spaces by IM James, and he defines the uniform subspace on page 97. Here is his definition:

Suppose now that we have a uniform space $$X$$ and a subset $$A$$ of $$X$$. For each subset $$D$$ of $$X \times X$$ the trace $$(A \times A) \cap D$$ on $$A \times A$$ is just the inverse image of $$D$$ under the injection $$A \times A \to X \times X$$. [...] We take the entourages of $$A$$ to be the traces on $$A \times A$$ of the entourages of $$X$$.

Let the uniformity on $$X$$ be denoted by $$\Phi$$. Am I right to understand this as saying that $$\Phi_A := \{\, U \subseteq A \times A \mid \exists V \in \Phi \text{ such that } U = V \cap (A \times A) \,\}$$ forms a uniformity on $$A$$?

Let me assume that the above was correct. I follow the definition of a uniformity on Wikipedia. I'm very uncertain of my proofs, especially (2) and (4). Here it is:

Observe that $$A \times A = (X \times X) \cap (A \times A)$$, so $$A \times A$$ belongs to $$\Phi_A$$ and $$\Phi_A$$ is non-empty.

1. If $$U \in \Phi_A$$, then $$\Delta A \subseteq U$$.

Let $$\Delta A = \Delta X \cap (A \times A)$$ be the diagonal of $$A \times A$$. Since $$\Delta X \subseteq V$$ for all $$V \in \Phi$$, it follows that $$\Delta A \subseteq U$$.

1. If $$U \in \Phi_A$$ and $$U \subseteq U' \subseteq A \times A$$, then $$U' \in \Phi_A$$.

Suppose $$U = V \cap (A \times A)$$ for some $$V \in \Phi$$, and $$U' = V' \cap (A \times A)$$ for some $$V' \subseteq X \times X$$. Define $$W = V \cup V' \subseteq X \times X$$. Then $$W$$ is in $$\Phi$$ because uniformities are upward-closed, and \begin{align*} W \cap (A \times A) &= (V \cap (A \times A)) \cup (V' \cap (A \times A)) \\ &= U \cup U' \\ &= U'. \end{align*} Therefore we have shown that $$U'$$ is in $$\Phi_A$$.

1. If $$U_1 \in \Phi_A$$ and $$U_2 \in \Phi_A$$, then $$U_1 \cap U_2 \in \Phi_A$$.

Suppose $$U_i = V_i \cap (A \times A)$$ for some $$V_i \in \Phi$$ for $$i = 1,2$$. Then we have $$U_1 \cap U_2 = (V_1 \cap V_2) \cap (A \times A)$$. Since $$\Phi$$ is closed under finite intersection, the set $$V_1 \cap V_2$$ belongs to $$\Phi$$. This implies that $$U_1 \cap U_2$$ is in $$\Phi_A$$.

1. If $$U \in \Phi_A$$, then there exists $$U' \in \Phi_A$$ such that $$U' \circ U' \subseteq U$$.

Fix $$U \in \Phi_A$$ and suppose $$U = V \cap (A \times A)$$ for some $$V \in \Phi$$, then there exists $$V' \in \Phi$$ such that $$V' \circ V' \subseteq V$$.

Let $$U' = V' \cap (A \times A)$$. We have \begin{align*} U' \circ U' &= \{\, (x,z) \in A \times A \mid \exists y \in A \text{ such that } (x,y), (y,z) \in U'\, \} \\ &\subseteq \{\, (x,z) \in X \times X \mid \exists y \in X \text{ such that } (x,y),(y,z) \in V'\, \} \cap (A \times A) \\ &= (V' \circ V') \cap (A \times A)\\ &\subseteq V \cap (A \times A) \\ &= U. \end{align*}

1. If $$U \in \Phi_A$$, then $$U^{-1} \in \Phi_A$$.

Suppose $$U = V \cap (A \times A)$$ for some $$V \in \Phi$$. Then we have \begin{align*} U^{-1} &= \{\, (y,x) \in A \times A \mid (x,y) \in U \, \}\\ &= \{\, (y,x) \in X \times X \mid (x,y) \in V\, \} \cap (A \times A)\\ &= V^{-1} \cap (A \times A). \end{align*} So $$U^{-1}$$ is in $$\Phi_A$$ because $$V^{-1}$$ is in $$\Phi$$.

Sorry for the long post. I'd really appreciate any comments/hints. Thanks!

In (2) you have to start with $$U \in \Phi_A$$ so $$U = V \cap (A \times A)$$ with $$V \in \Phi_X$$. Then assume $$U \subseteq U' \subseteq A \times A$$ and note that
$$(V \cup U') \cap (A \times A) = (V \cap (A \times A)) \cup (U' \cap (A \times A)) = (V \cap (A \times A)) \cup U' = U \cup U' =U'$$ and $$V \cup A' \in \Phi_X$$ by upwards closedness. And thus also $$U' \in \Phi_A$$.
So I did not assume $$U'$$ was already a subset of $$X \times X$$ intersected with $$A \times A$$, but simply showed it to be of this form.
(4) is essentially fine. If $$U \in \Phi_A$$ is of the form $$V \cap (A \times A)$$ and we have $$V' \in \Phi_X$$ with $$V' \circ V' \subseteq V$$ then indeed $$U':= V' \cap (A \times A)\in \Phi_A$$ is as required, as $$U' \circ U' = (V' \cap (A \times A)) \circ (V' \cap (A \times A))= (V' \circ V') \cap (A \times A) \subseteq V \cap (A \times A) = U$$ where the first distributivity between $$\circ$$ and $$\cap$$ is a general fact of relations and easy to see by definition analysis.
• Thanks for the detailed response! About (2), why can't we assume that $U' \subseteq A \times A$ is already a subset of $X \times X$? Does it not follow from $A$ being a subset of $X$? – jessica Sep 24 '18 at 4:29
• @jessica I mean that a priori we don’t assume that $U’$ is of the form $V’\cap(A \times A)$, because that’s a part of what we have to show to see that $U’ \in \Phi_A$. – Henno Brandsma Sep 24 '18 at 4:36