Metric Space, Show that it's open

szklany
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Let (X,d) be a metric space, and x is an element in X. Show that \{y \in X|d(y,x)>r\} is open for all r in Reals.

I really need some help with this one, I have almost no idea on how I am meant to solve this.

The only thing i know is that I have to use the Openness definition, that states something like \forall x_0 \in U \exists r>0| B_r \in U, where in U is a subelement of the metric space (X,d).

But i don't know how to get started.
 
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szklany said:
Let (X,d) be a metric space, and x is an element in X. Show that \{y \in X|d(y,x)>r\} is open for all r in Reals.

I really need some help with this one, I have almost no idea on how I am meant to solve this.

The only thing i know is that I have to use the Openness definition, that states something like \forall x_0 \in U \exists r>0| B_r \in U, where in U is a subelement of the metric space (X,d).

But i don't know how to get started.
Let a be a point in {y| d(x,y)> r}. Then d(x,a)> r. Construct the neighborhood about a with radius (d(x,a)- r)/2. If b is any point in that neighborhood, use the triangle inequality to show that d(x, b)> r also.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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