1. ## Continuity on M

Fix a point a in a metric space (M,d), and define f:M->R by f(x)=d(x,a). Show f is continuous on M.

I took a sequence Xn->x in (M,d)
So f(Xn)=d(Xn,a)
taking the limit, f(Xn)->d(x,a)=f(x)
Since we have Xn->x => f(Xn)->f(x), f is continuous on M.

But I feel like that's wrong, because how do we know f(x) is in R?
I need to show if Xn->x in (M,d), then f(Xn)->f(x) in R.....but I don't know whether or not f(x) is in R do I? Am i missing something obvious?
Thanks guys.

2. f(x) is in R because f(x) is a distance, whowever I think you nedd to prove d(xn,a)->d(x,a)

3. Originally Posted by cp05
Fix a point a in a metric space (M,d), and define f:M->R by f(x)=d(x,a). Show f is continuous on M.

I took a sequence Xn->x in (M,d)
So f(Xn)=d(Xn,a)
taking the limit, f(Xn)->d(x,a)=f(x)
Since we have Xn->x => f(Xn)->f(x), f is continuous on M.

But I feel like that's wrong, because how do we know f(x) is in R?
I need to show if Xn->x in (M,d), then f(Xn)->f(x) in R.....but I don't know whether or not f(x) is in R do I? Am i missing something obvious?
Thanks guys.
Why not use the normal definition of continuity? It's easier.

Then, $d(x,a)-d(x_0,a)\leqslant d(x,x_0)+d(x_0,a)-d(x_0,a)=d(x,x_0)$ and $d(x_0,a)-d(x,a)\leqslant d(x_0,x)+d(x,a)-d(x,a)=d(x_0,x)$. Combining these two we see that $\left|d(x,a)-d(x_0,a)\right|\leqslant d(x,x_0)$ and thus $f$ is Lipschitz and thus trivially continuous.

4. I'm not familiar with Lipschitz What is that?

5. Originally Posted by cp05
I'm not familiar with Lipschitz What is that?
If $\varphi:M\to M'$ is a function (where $M,M'$ are metric spaces) then $\varphi$ is said to be Lipschitz with Lipschitz constant $C>0$ if $d_{M'}(\varphi(x),\varphi(y))\leqslant C d_M(x,y)$. I actually didn't prove Lipschitz per se (since I fixed $x_0$) but it can easily be proved using the EXACT same idea that $f$ is Lipschitz.

Every Lipschitz function is uniformly continuous and thus continuous. It's clear for if $\varepsilon>0$ is given then choosing $\delta=\frac{\varepsilon}{C}$ we see that $d_M(x,y)<\delta\implies d_{M'}(\varphi(x),\varphi(y))\leqslant C d_M(x,y)\leqslant C\frac{\varepsilon}{C}=\varepsilon$