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→Two-dimensional case: The case explained is not one dimensional! It is two dimensional, x and y axes are even shown in the graph. Tags: Reverted Mobile edit Mobile web edit |
Undid revision 1295555632 by 2A02:2149:8626:7200:59:1FDE:4418:8190 (talk) it's the case for the unit interval, which is 1-dim'l; it's just the graph being drawn in 2 |
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Brouwer "flattens" his sheet as with a flat iron, without removing the folds and wrinkles. Unlike the coffee cup example, the crumpled paper example also demonstrates that more than one fixed point may exist. This distinguishes Brouwer's result from other fixed-point theorems, such as [[Stefan Banach]]'s, that guarantee uniqueness.
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[[File:Théorème-de-Brouwer-dim-1.svg|200px|right]]
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Intuitively, any continuous line from the left edge of the square to the right edge must necessarily intersect the green diagonal. To prove this, consider the function ''g'' which maps ''x'' to ''f''(''x'') − ''x''. It is ≥ 0 on ''a'' and ≤ 0 on ''b''. By the [[intermediate value theorem]], ''g'' has a [[Root of a function|zero]] in [''a'', ''b'']; this zero is a fixed point.
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