"There's a subsequence of such triangles such that the three corners all converge to the same point z" may not be true and you don't need it, right? You only need there's a subsequence of red points converging to z, a subsequence of blue points converging to z and a subsequence of green points converging to z. But these three subsequences may come from different triangles.
A parallel line to h(x)=x would have to be either completely above x or completely below x. If it is completely above, the function value at 1 would be strictly larger than 1 and this is not allowed. The function has to map from [0,1] to [0,1]. If it is completely below, the function value at 0 would be strictly smaller than 0 and this is also not allowed.
Excellent explanations. Thank you very much.
At 05:20 it should be red if f(x) >= x and blue if f(x)
"There's a subsequence of such triangles such that the three corners all converge to the same point z" may not be true and you don't need it, right? You only need there's a subsequence of red points converging to z, a subsequence of blue points converging to z and a subsequence of green points converging to z. But these three subsequences may come from different triangles.
Without the triangle things, how would you argue there’s a subsequence of red, green, blue points that converge to the same point z?
3:10 what if f(x) is parallel to h(x)=x then there will never be an f(x)=x. What is the error in my thinking?
A parallel line to h(x)=x would have to be either completely above x or completely below x. If it is completely above, the function value at 1 would be strictly larger than 1 and this is not allowed. The function has to map from [0,1] to [0,1]. If it is completely below, the function value at 0 would be strictly smaller than 0 and this is also not allowed.
Thank you Sir. Which device is used to prepare the class.
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