Real-life recording of a University of Oxford Mathematics tutorial with Dr Tom Crawford. The lesson goes over a past exam paper as part of a revision tutorial covering the second year undergraduate Metric Spaces and Complex Analysis course.
Real-life recording of a University of Oxford Mathematics tutorial with Dr Tom Crawford. The lesson goes over a past exam paper as part of a revision tutorial covering the second year undergraduate Metric Spaces and Complex Analysis course.
TOM ROCKS MATHS 
Hi,
I found the very (too) simple observation…
Assume x2 != x1 being two distinct fixed points of F(x)
We have that F satisfies d(F(x1), F(x2)) < d(x1, x2) (1)
x1 and x2 being fixed points we also have d(F(x1), F(x2)) = d(x1, x2) (2)
(1) and (2) are contradictions therefore it is not possible for F(x) to have two distinct fixed points
Is this way too simple ? I made no use of compactness or completeness either …
Thank you
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In my previous comment, the demonstration I made was for the uniqueness of the fixed point…
However,
I just realized the question was also to demonstrate the existence of such a fixed point…
and in that case, my idea would be to begin with 2 distinct points and iteratively apply F(x) to each points, from d(F(x1), F(x2)) < d(x1, x2) we get a sequence that converges and with completeness the limit exist and is likely as close as we need to the fixed point, bracketed by x1_n and x2_n with x1_(n+1) = F(x1_n)… same for x2_n
That would be my main idea…
Note, I am 62yo got my BSc in 1984 (a long time ago š and currently retired … I just got interested by the video and wrote faster than my common sense to stop me…
Thank you again for such fantastic work you do …
Catherine
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