Recent content by dillingertaco

We only used properties available from the metric. For it to work with every metric you could say this: by the triangle inequality on (X,d) (from my earlier post) we have: [itex]d_{R}(d_{(X,d)}(x_{n},y_{n}),d(x,y)) \leq d_{R}(d_{(X,d)}(x_{n},x)+d_{(X,d)}(x,y)+d_{(X,d)}(y,y_{n}),d(x,y))[\itex]...

It looks like your confusing is coming from defining it as 2|x-y| and then the 2 not being on what we did. I was thinking of d(x,y)=|x-y| and there was no 2. Either way even if there was a 2 you could choose epsilon such that it doesn't matter.

Where did that plus sign come from?

I think there is a better way to formalize it, which I've been thinking about. The only thing you know about metrics is they must follow the triangle inequality So you know by the triangle ineq: d(x_{n},y_{n})\leq d(x{n},x)+d(x,y_{n})\leq d(x{n},x)+d(x,y)+d(y,y_{n}) So from there assume n is...

I added more to my post. All metrics map into \mathbb{R}. So therefore d(x,y) is just a real number. So you can use |d(x_n,y_n)-d(x,y)|. Yes, |A-B| is the standard metric on R, and this fits into that form. In this case |A-B| simple means absolute value. Ex: |3-5|=2. EDIT: Absolute value in...

I think what you're missing is a metric is a function into R. So d(x,y) is a scalar and so is the other one. So in that case you can represent it like this: |d(x_{n},y_{n})-d(x,y)| I'm not sure about this part, but someone can elaborate. If you treat the metric as a continuous function, we...

<A+B,A+B>=<A-B,A-B> By properties of the dot product.. <A,A>+2<B,A>+<B,B>=<A,A>-2<B,A>+<B,B> Get everything to one side and deduce from that.

I think the way I'm thinking of it defining x=min(d/2,1). That would make me happy. Thank you so much for your help.

That's a better idea. Then |\frac{\delta}{2}^{2}-4|\geq 1 since \delta/2\leq1/2 Would there be a way to do this without the caveat when \delta>1. it seems very inelegant.

Hello micromass, I appreciate the response! I'm getting confused here how to formalize this. So for all \delta, we can find such an x as you described. It makes sense in the terms of delta = 1, look at x= 1/2, with \epsilon = 1. I see that this case works, but how do do this for every delta...

Homework Statement Prove the function f(x)= { 4 if x=0; x^2 otherwise is discontinuous at 0 using epsilon delta. Homework Equations definiton of discontinuity in this case: there exists an e>0 such that for all d>0 if |x-0|<d, |x^2-4|>e The Attempt at a Solution I'm confused...