# 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 $\u03f5$ be the proximities generated by (http://planetmath.org/UniformProximity) $\mathcal{U}$ and $\mathcal{V}$ respectively. It is known that $X$ and $Y$ are proximity spaces with proximities $\delta $ and $\u03f5$ 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 $f(A)\u03f5f(B)$, or equivalently,

$$V[f(A)]\cap V[f(B)]\ne \mathrm{\varnothing},$$ |

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 $U[A]\cap U[B]\ne \mathrm{\varnothing}$, 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

$$V[f(A)]\cap V[f(B)]\ne \mathrm{\varnothing}.$$ |

This shows that $f$ is proximity continuous. ∎

Title | proof that uniformly continuous is proximity continuous |
---|---|

Canonical name | ProofThatUniformlyContinuousIsProximityContinuous |

Date of creation | 2013-03-22 18:07:50 |

Last modified on | 2013-03-22 18:07:50 |

Owner | CWoo (3771) |

Last modified by | CWoo (3771) |

Numerical id | 6 |

Author | CWoo (3771) |

Entry type | Proof |

Classification | msc 54E15 |

Classification | msc 54C08 |

Classification | msc 54C05 |

Classification | msc 54E17 |

Classification | msc 54E05 |