Thread: question in metric topology

Let (X,d) be a metric space, Given a closed subset Y of X and a point x in X. Show that d(x,Y) is continuous as a function of x. that is f(x)=d(x,Y) is continuous

Show also that a topology induced by the metric d is finer than the cofinite topology on X.

Thank you very much for any help or guidance

Originally Posted by fuzzy topology

Let (X,d) be a metric space, Given a closed subset Y of X and a point x in X. Show that d(x,Y) is continuous as a function of x. that is f(x)=d(x,Y) is continuous

Show also that a topology induced by the metric d is finer than the cofinite topology on X.

Thank you very much for any help or guidance

The first one is just messing around with the triangle inequality, what have you tried? For the second one let $\displaystyle \mathcal{T}_C$ be the cofinite topology on $\displaystyle X$ and $\displaystyle \mathcal{T}_C$ the metric. Note then that if $\displaystyle U\in\mathcal{T}_M$ then $\displaystyle \#\left(X-U\right)<\infty$ and thus $\displaystyle X-U$ is in $\displaystyle \left\{X-O:O\in\mathcal{T}_M\right\}$ since any metric space is $\displaystyle T_1$ and thus $\displaystyle U\in\mathcal{T}_M$.


(just for the record, the cofinite topology on a space is the coarsest $\displaystyle T_1$ topology)

For the first question, I tried the following
given epsilon > 0 , there is delta > 0, such that if d(x,y)<delta, then d(f(x),f(y)) should be less than epsilon, that is what we want.
But d(x,y) < d(x,Y)+d(y,Y) =f(x)+f(y)

Unfortunately I can not reach to the desired goal. Is what I wrote true. Please help me and thank you very very much for helping in solving the second question

This is the theorem that you must prove.
If $\displaystyle A$ is a set and $\displaystyle p~\&~q$ are points then $\displaystyle \left| {D(A;p) - D(A;q)} \right| \leqslant d(p,q)$.
Hints: Show that $\displaystyle \left( {\forall a \in A} \right)\left[ {\text{glb} \left\{ {d(x,a):x \in A} \right\} \leqslant d(p,a)} \right]$ then
do it for $\displaystyle q$ and observe $\displaystyle D(A,p)-D(A,q)\le d(p,q)$.