Let $(X,d)$ be a metric space and $ a \in X $
Show that $f(x) = d(x,a)$ is a Lipschitz function from X to $\Re$
Use this fact to show $S$ is an open subset of $X$:
Let $(X,d)$ be a metric space and $a \in X, r > 0.$ Let $S$={$x \in X : d(a,x) > r$}
>Lipschitz Defn:
Let $f$ be a function from $(X,d)$ to $(Y,\rho)$
$f$ is said to be Lipschitz if:
$\exists$ K $\geq$ 0 s.t. $\rho( f(x_1) , f(x_2) ) \leq$ Kd($x_1$,$x_2$) $\forall$ $x_1 x_2 \in X$
>Continuity Defn:
Let $(X,d)$ and $(Y,\rho)$ be metric spaces, and $f: X \to Y$
Then $f$ is continuous $iff$ f$^{-1}$($B$) = { $x \in X : f(x) \in B $} is an open subset of $X$, whenever $B$ is an open subset of $Y$
>Triangle Inequality for a Metric:
$d(x,z) \leq d(x,y) + d(y,z)$
My first question is, am I thinking about this correctly?
If $f(x)$ is a Lipschitz function, it is uniformly continuous
If $f(x)$ is uniformly continuous, it is continuous at every point
If $f(x)$ is continuous at every point $\Rightarrow$ the inverse images of open sets are open
Hence $S$ is an open subset of $X$.
Secondly, how would you formally construct this?
We know $f(x) = d(x,a)$ s.t. $f:X \to \Re$
By definition of Lipschitz we have $\exists K \geq 0$ s.t. $\rho(f(x_1),f(x_2)) \leq Kd(x_1,x_2) \forall x_1,x_2 \in X$
& since our codomain is $\Re$
$\rho(f(x_1),f(x_2)) = | d(x_1,a) - d(x_2,a) | \leq Kd(x_1,x_2)$
Taking $K=1$
$\ldots$
Help in finishing this off would be appreciated.
Thanks.