Consider the two metrics and
(you may assume they are metrics).
i) Show that d and δ are not equivalent.
Last edited by QuantumRoulette; 7 Mar 2017 at 10:30 PM. Reason: fixed Tex issues
I assume d(x,y) is supposed to be ||x-y|| and the set is some normed vector space V with norm ||.|| (e.g. R^d). (Please specify more if this is not the intended setting.)
Then these two metrics are not (strongly) equivalent because V is bounded with the delta metric but unbounded with the d metric.
That V is bounded with the delta metric follows immediately from the definition of delta, which must always lie in [0,1). On the other hand d(tx,0)=|t|d(x,0) can be made arbitrarily large for nonzero x.
Note that these two metrics ARE topologically equivalent though, in the sense that convergence in one metric implies convergence in the other. This follows from from the map x->x/(1+x) being a homeomorphism from [0,inf) to [0,1).
Might not be easy to visualise the graph of the function on all of R^2, but you should certainly be able to visualise what it looks like on the slices x=const. or y=const which is all that matters for seeing/proving the nonexistence of the iterated limit. It is an oscillatory expression that oscillates faster as you approach axes. One of the two summands becomes irrelevant as you get close to the axes, so the other one dominates. This thing behaves like (const).sin(1/x), which of course does not converge unless that const is zero.
The boundedness of sine makes it clear that f(x,y) tends to zero as (x,y) tends to zero though.
Long story short: don't be too hasty to form intuitions in analysis, lots of things can behave weirdly...you kind of have to slowly build up a list of things that ARE true (via proof!) rather than assuming innocuous statements are true and ruling these things out as you come across pathological counterexamples.
And when you are looking at functions like this, try to isolate the terms that actually matter for the property you are trying to prove. A large part of analysis is just approximating ugly things by nice things, throwing away small sets on which a function behaves badly, etc etc.
There was a typo; it was meant to be a less than or equals to. I think InteGrand just ignored or figured I had typo'd and just used the correct inequality.
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As in, for every ball around 0 there's always an irrational number in it, therefore 0 is not an interior point so it cannot be open. (I hope I did not screw this up.)
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I wanted to use the floor/ceiling functions but then I realised that'd be problematic if r = 0.1, so what might be a good choice for a rational number that is in every ball B(sqrt2, r)?
(Sorry, I think I worded my question horribly)
Last edited by leehuan; 11 Mar 2017 at 11:26 AM.
That is one beautiful trick.
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I definitely don't want to use the Lagrange multiplier theorem to find the closest point (x,y) to work off, so any pointers on what radius I should choose my ball to be? (to prove an arbitrary point (x,y) is interior)
Mate, stop cheating on these assignment questions. Do it yourself ffs.
Just a brief sketch-out please
I know that f is not continuous at 0 but I'm not sure if that helps since we're talking about partial derivatives here.
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