Contraction Mappings without a fixed point

Show that the following mapping $\displaystyle f:X \rightarrow X$ do not have a fixed point and explain why the Contraction Mapping Principle is not contradicted.
Problem: Let X=R and f(x) = x+1 for all x in X.

My Solution:
f do not have a fixed point because x does not equal to x+1 for any real number x.

d(f(x),f(y)) = |(x+1) - (y+1)| = |x-y|,
and d(x,y) = |x-y|.

So d(f(x),f(y)) = d(x,y), now I know I have to find a constant c with 0 <= C < 1 for the Contraction Mapping Principle to hold. But I can't find that c.

Please help.

Quote:

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Originally Posted by tttcomrader

Problem: Let X=R and f(x) = x+1 for all x in X.

Show that the mapping doesn't have a fixed point, and the Contraction Mapping Theorem is not contradicted.

Solution: d(f(x),f(y)) = |(x+1) - (y+1)| = |x-y|,
and d(x,y) = |x-y|.

So d(f(x),f(y)) = d(x,y), now I know I have to find a constant c with 0 <= C < 1 for the Contraction Mapping Principle to hold. But I can't find that c.

Please help.

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Look at the question and answer what is asked.

1. Does the mapping have a fixed point? Justify your answer.

2. If there was no fixed point show that at least one of the conditions
of the the contraction mapping theorem is violated.

RonL

I think I didn't word the question correctly, let me edit.

"Show that the following mapping $\displaystyle f:X \rightarrow X$ do not have a fixed point and explain why the Contraction Mapping Principle is not contradicted."

So if I understand this question correctly, I have to show the mapping do not have a fixed point, but yet it is still a contraction. (I guess because it is not in a complete space?)

And yeah, I forgot to put the first part of my answer on there, it is added now.

Quote:

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Originally Posted by tttcomrader

I think I didn't word the question correctly, let me edit.

"Show that the following mapping $\displaystyle f:X \rightarrow X$ do not have a fixed point and explain why the Contraction Mapping Principle is not contradicted."

So if I understand this question correctly, I have to show the mapping do not have a fixed point, but yet it is still a contraction. (I guess because it is not in a complete space?)

And yeah, I forgot to put the first part of my answer on there, it is added now.

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No, "the Contraction Mapping Principle is not contradicted." You need to show why the mapping does not have a fixed point and why there is no contradiction.

-Dan

So what I need is to find a c with 0 <= c < 1 such that d(f(x),f(y)) <= cd(x,y) right?

Now as how my calculation goes, d(f(x),f(y)) = d(x,y), so I"m stuck here. Or perhaps I have done the whole thing wrong.