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Is there any two uniformities $\mathcal{D}_1$ and $\mathcal{D}_2$ on a set $X$ such that $$\mathcal{D}_1\cap \mathcal{D}_2$$ is not a uniformity on $X$?

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up vote 4 down vote accepted

HINT: Yes, such examples exist. I’m assuming that these are diagonal uniformities. Let $X=[0,1]$, and let $\Delta$ be the diagonal in $X\times X$. Let $a,b\in(0,1)$ with $a\ne b$, and let

$$\mathscr{D}_1=\big\{D\subseteq X\times X:\Delta\cup\{\langle a,1\rangle,\langle 1,a\rangle\}\subseteq D\big\}$$


$$\mathscr{D}_2=\big\{D\subseteq X\times X:\Delta\cup\{\langle b,1\rangle,\langle 1,b\rangle\}\subseteq D\big\}\;;$$

show that these are uniformities on $X$. Let $\mathscr{A}=\mathscr{D}_1\cap\mathscr{D}_2$; clearly

$$\mathscr{A}=\{A\subseteq X\times X:A_0\subseteq A\}\;,$$

where $A_0=\Delta\cup\{\langle a,1\rangle,\langle 1,a\rangle,\langle b,1\rangle,\langle 1,b\rangle\}$. Each $A\in\mathscr{A}$ is a relation on $X$, so $A\circ A$, the composition of $A$ with itself, makes sense. Show that $\langle a,b\rangle\in A\circ A$ for all $A\in\mathscr{A}$. Conclude that for all $A\in\mathscr{A}$, $A\circ A\nsubseteq A_0$, so $\mathscr{A}$ cannot be a uniformity on $X$. (Specifically, it violates condition (4) of this definition of diagonal uniformity, which can be rephrased in terms of composition as saying that for each $U\in\Phi$ there is a $V\in\Phi$ such that $V\circ V\subseteq U$.)

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