Uni Stuff > Mathematics

Analysis

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kamil9876:
Be more specific: x=0? y=0? a single number is not what you're interested in in this space. (I'm assuming the space is )

Anyways, if it helps: as x approaches 0 from the positive side, then the point (x,y) does not approach any point.

/0:
Oh, thanks kamil, sorry I meant and .

So because there is not a unique limit point as or then the set is closed because all of its limit points are contained in the set?

humph:

--- Quote from: /0 on April 26, 2010, 10:55:59 pm ---Oh, thanks kamil, sorry I meant and .

So because there is not a unique limit point as or then the set is closed because all of its limit points are contained in the set?

--- End quote ---
Exactly. If were a limit point, then there'd be a sequence approaching it, which clearly there's not.
P.S. you could just think of it as the graph of , which is the union of two curves in .

kamil9876:
Yes it does contain all it's limit points, so it is closed.

As for your fear about x approaching 0: if you consider any sequence that approaches 0, like for example Then the sequence of the points (where ) does not approach any point in the space (think :the distance between and some fixed point in space approach infinity as approaches infinity so it doesn't converge to anything) hence no need to worry about such a sequence when checking for closedness.

/0:
Thanks :)

Show using the contraction mapping theorem that



has a fixed point.

Whenever we aren't told what the metric is, should we automatically assume the euclidean metric?

Anyway, I did




In the working they say this is

I'm not really sure how this happens...

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