Can someone help me with this? it seems really easy question but i couldn't see it through...
For a compact Subset $K$ of a metric space $X$ and $x \in X$. The Function is $f:X\to R$ given by $f(x) = d(x,K)$.
Show that $f$ is continuous.
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Let $x \in X$, let's show that f is continuous in $x$. Let $t \in X$ and $\epsilon > 0$ By definition of $d$, $\exists a \in K$ such as, $d(x,a) \le d(x,K) + \epsilon$. So $d(t,K) \le d(t,x) + d(x,a) \le d(t,x) + d(x,K) + \epsilon$. So $f(t) \le d(t,x) + f(x) + \epsilon$, if we make $\epsilon$ tends to $0$, we have $f(t) \le f(x) + d(t,x)$.
With the same reasonning we have $f(x) \le f(t) + d(t,x)$. So $|f(x)-f(t)| \le d(x,t)$, so $f$ is 1-lipschitz, so $f$ is continuous.
Bérénice
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i almost got it but how do you say $$ d(x,a) \le d(x,K) + e $$ – suugii19 Apr 09 '16 at 11:38
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$d(x,K)=inf(d(x,k)|k \in K)$, so you you obtain the inegality by definition of the infinum. So if you fix the $\epsilon$, you can find $a \in K$ such as $d(x,a)$ is small enough to fit in the inequality. – Bérénice Apr 09 '16 at 11:43
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Now i get it. thank you Jennifer! :) – suugii19 Apr 09 '16 at 11:52