rayquaza17
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ImageI don't even know how to start.
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Noble.
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This is essentially an extension of the usual triangle inequality. This is one of the few times I advocate thinking about things geometrically (:lol:)

You can think of the LHS as a 'straight line' from x to y

The RHS is the journey: x to a, a to b, b to y.

The latter is only ever going to be as short as the straight line x to y if a and b lie on the straight line from x to y.

(Note, the algebraic proof to this is also straight forward, but thinking about it geometrically can help motivate how to start)
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Noble.
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If that isn't helpful, the key hint is to use the triangle inequality.
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physicsmaths
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(Original post by Noble.)
If that isn't helpful, the key hint is to use the triangle inequality.
Is it possible with vectors?


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TeeEm
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(Original post by rayquaza17)
ImageI don't even know how to start.
|x-y|=|x-a+a-b+b-y|=|(x-a) + (a-b) + (b-y)| <= .... by a famous result
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rayquaza17
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(Original post by TeeEm)
|x-y|=|x-a+a-b+b-y|=|(x-a) + (a-b) + (b-y)| <= .... by a famous result
What's the famous result?

Thanks, I had written down (x+a)+(b-a)-(y+b) [after seeing Noble's hint] and I was wondering why it didn't work.

Edit: I've now realised I had it written down like that because I had written the question down wrong.:facepalm:

(Original post by Noble.)
If that isn't helpful, the key hint is to use the triangle inequality.
Nope, it was helpful! Thank you!

I know it was such a simple question, but it's one of the first ones on a long list of analysis questions I need to do and I think because it's on that list that my mind is going blank. (Analysis and me don't really get on.)
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Noble.
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(Original post by physicsmaths)
Is it possible with vectors?


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Yes, you can do it with \mathbb{R}^n as well, from the same geometric principal.
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Noble.
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(Original post by rayquaza17)
What's the famous result?

Thanks, I had written down (x+a)+(b-a)-(y+b) [after seeing Noble's hint] and I was wondering why it didn't work.



Nope, it was helpful! Thank you!

I know it was such a simple question, but it's one of the first ones on a long list of analysis questions I need to do and I think because it's on that list that my mind is going blank. (Analysis and me don't really get on.)
You can do it his way, or you can approach it from the other end

|x-a| + |a-b| + |b-y| \geq |x-b| + |b-y| \geq |x-y|

With the first step applying the triangle inequality to the first two terms, then the second step applying it again.
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rayquaza17
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Ok this is a different question, but I think it's the same sort of method:
Image

Would this be right:
x_n converges with limit l means, given any \frac{\epsilon }{2}&gt;0, there is N\in \mathbb{N} such that |x_n-l|&lt;\frac{\epsilon }{2} for all n\geq N

Then we can write |l-x_m|&lt;\frac{\epsilon }{2}

Then:
|x_n-x_m|=|x_n-l+l-x_m|\leq |x_n-l|+|x-x_m|&lt;\frac{\epsilon }{2}+\frac{\epsilon }{2}=\epsilon
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Noble.
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Yes. What you've shown is that a convergent sequence is Cauchy. Just to nitpick, the question should say |x_n - x_m| &lt; \epsilon for all n,m \geq N, not just n.

You can try the harder direction as well, that every Cauchy sequence is convergent.
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