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- #1

- Jan 29, 2012

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Prove that a contraction T on a metric space is a continuous mapping? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

- Thread starter Fernando Revilla
- Start date

- Thread starter
- #1

- Jan 29, 2012

- 661

Prove that a contraction T on a metric space is a continuous mapping? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

- Thread starter
- #2

- Jan 29, 2012

- 661

The result is more general. Suppose $(E,d)$ is a metric space and $T:E\to E$ is a Lipschitz mapping, i.e. there is a positive constant $K$ such that $d(T(x),T(y))\le Kd(x,y)$ for all $x,y\in E$. Then, for all $\epsilon >0$ and choosing $\delta=\epsilon/K:$ $$d(x,y)<\delta \Rightarrow d(T(x),T(y))\le Kd(x,y)<K\frac{\epsilon}{K}\Rightarrow d(T(x),T(y))<\epsilon$$ This means that every Lipschitz mapping is uniformly continuous, as a consequence continuous. But a contraction map is a Lipschitz mapping with $K<1$, hence every contraction is a continuous mapping.