Let A be a non-empty set in a metric space (X,d). Define$\displaystyle f : X \rightarrow \mathbb{R}$ by $\displaystyle f(x) = inf \{d(a,x) : a \in A \}$

. Prove that f is continuous.

I dont see how to bound f(x) and f(y)

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- Dec 30th 2011, 02:31 AMFGT12Proving a function is continuous in a metric spaces
Let A be a non-empty set in a metric space (X,d). Define$\displaystyle f : X \rightarrow \mathbb{R}$ by $\displaystyle f(x) = inf \{d(a,x) : a \in A \}$

. Prove that f is continuous.

I dont see how to bound f(x) and f(y) - Dec 30th 2011, 03:10 AMFernandoRevillaRe: Proving a function is continuous in a metric spaces
The class of subsets of $\displaystyle \mathbb{R}$ in the form $\displaystyle G_{\alpha}=(\alpha,+\infty)$ or $\displaystyle H_{\beta}=(-\infty,\beta)$ is a subbasis of $\displaystyle (\mathbb{R},T_u)$ . Prove that $\displaystyle f^{-1}(G_{\alpha})$ and $\displaystyle f^{-1}(H_{\beta})$ are open sets.*Hint* - Dec 30th 2011, 03:23 AMPlatoRe: Proving a function is continuous in a metric spaces
Suppose that $\displaystyle a\in A$.

$\displaystyle f(x)=\inf\{d(x,t):t\in A\}\le d(x,a)$

$\displaystyle d(a,x)\le d(a,y)+d(y,x)$ or $\displaystyle d(a,x)-d(y,x)\le d(a,y)$ for each $\displaystyle a\in A$

Thus $\displaystyle f(x)-d(x,y)\le f(y)$ or $\displaystyle f(x)-f(y)\le d(x,y)$

Can you continue to prove that $\displaystyle |f(x)-f(y)|\le d(x,y)~?$ - Dec 30th 2011, 04:27 AMILikeSerenaRe: Proving a function is continuous in a metric spaces
- Dec 30th 2011, 05:02 AMPlatoRe: Proving a function is continuous in a metric spaces
- Dec 30th 2011, 05:26 AMILikeSerenaRe: Proving a function is continuous in a metric spaces