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# proof that uniformly continuous is proximity continuous

Let $f:X\to Y$ be a uniformly continuous function from uniform spaces $X$ to $Y$ with uniformities $\mathcal{U}$ and $\mathcal{V}$ respectively. Let $\delta$ and $\epsilon$ be the proximities generated by $\mathcal{U}$ and $\mathcal{V}$ respectively. It is known that $X$ and $Y$ are proximity spaces with proximities $\delta$ and $\epsilon$ respectively. Furthermore, we have the following:

###### Theorem 1.

$f:X\to Y$ is proximity continuous.

###### Proof.

Let $A,B$ be any subsets of $X$ with $A\delta B$. We want to show that $f(A)\epsilon f(B)$, or equivalently,

$V[f(A)]\cap V[f(B)]\neq\varnothing,$ |

for any $V\in\mathcal{V}$. Pick any $V\in\mathcal{V}$. Since $f$ is uniformly continuous, there is $U\in\mathcal{U}$ such that

$U[x]\subseteq f^{{-1}}(V[f(x)]),$ |

for any $x\in X$. As a result,

$U[A]\subseteq f^{{-1}}(V[f(A)]),$ |

which implies that

$f(U[A])\subseteq V[f(A)].$ |

Similarly $f(U[B])\subseteq V[f(B)]$. Now, $A\delta B$ is equivalent to $U[A]\cap U[B]\neq\varnothing$, so we can pick

$z\in U[A]\cap U[B].$ |

Then

$f(z)\in f(U[A])\cap f(U[B])\subseteq V[f(A)]\cap V[f(B)],$ |

and therefore

$V[f(A)]\cap V[f(B)]\neq\varnothing.$ |

This shows that $f$ is proximity continuous. ∎

## Mathematics Subject Classification

54E15*no label found*54C08

*no label found*54C05

*no label found*54E17

*no label found*54E05

*no label found*

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