# Continuity in Metric Spaces

#### I-Think

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