Continuity in Metric Spaces

Apr 2009
288
56
Just need to check the details in my proof are correct
Question
\(\displaystyle (X,d)\) is a metric space, fix \(\displaystyle x_0\in{X}\), let \(\displaystyle F_{x_0}=d(x_0,x)\). Show this function is continuous

Proof
Consider the open set \(\displaystyle W\in{\mathbb{R}}\) and let
\(\displaystyle F_{x_0}^{-1}(W)=[x\in{X}|F_{x_0}(x)=d(x_0,x)\in{W}}\)]. For all \(\displaystyle y\in{W}\), \(\displaystyle \exists{\epsilon_y}>0\) such that \(\displaystyle B(y,\epsilon_y)\subset(W)\)
Let \(\displaystyle z=F_{x_0}^{-1}\) so \(\displaystyle z\in{F_{x_0}^{-1}}\), consider \(\displaystyle B(z,\epsilon_y)\)

Let \(\displaystyle \alpha{\in{B(z,\epsilon_y)}\),\(\displaystyle F_{x_0}^{-1}(\alpha)\)\(\displaystyle =d(x_0,\alpha)<d(x_0,z)+d(z,\alpha)=y+\epsilon_y\)
So \(\displaystyle F_{x_0}^{-1}(\alpha)\in{B(y,\epsilon_y)}\subset{W}\), so \(\displaystyle B(z,\epsilon_y)\subset{F_{x_0}^{-1}(W)}\)
So \(\displaystyle {F_{x_0}^{-1}(W)\) is open and the result follows