# Fixed point theorem for the strict contraction

I want to prove the following fixed point problem:

Let $$(X,d)$$ be an arbitrary metric space and $$T:X\rightarrow X$$ a map satisfying $$d(Tx,Ty) whenever $$x\neq y.$$ Assume that for some $$x\in X$$, the sequence $$\{T^nx\}$$ has a subsequence converging to a point $$u$$. Prove that $$u$$ is a fixed point for $$T$$.

Can someone give a comment on my proof?. Here is my attempt:

Since $$\displaystyle\lim_{k\rightarrow\infty}T^{n_k}x=u.$$ Given $$\varepsilon>0$$, there exists $$N\in\mathbb{N}$$ such that for all $$k\geq N$$, $$d(u,T^{n_k}x)<\varepsilon.$$ Let $$m=n_k+l$$ be any positive integer which is greater $$n_k$$. Then $$d(u,T^{m}x)=d(T^lu,T^{n_k+l}x) This shows that $$T^mx\rightarrow u.$$ $$T$$ is continuous which is clear from the strict contraction. Thus the continuity of $$T$$ implies $$T(T^mx)\rightarrow Tu$$ and $$\{T^{m+1}x\}$$ is a subsequence of $$T^mx$$. We deduce $$Tu=u$$ and $$u$$ is a fixed point for $$T$$.

• Why is $d(u,T^mx)=d(T^lu, T^{n_k+l}x)?$ – Adrian Keister Jun 27 at 18:43
• Well, I'm thinking that $T$ is continuous and so $\lim_{k\rightarrow\infty}T^{n_k+l}=T^l(\lim_{k\rightarrow\infty}T^{n_k}x)=T^lu$. On the other hand, $\{T^{m}x\}$ can be viewed as a subsequence of $\{T^{n_k}x\}$. Consequently, we get $d(u,T^mx)=d(u,T^{n_k}x)$ and $d(u,T^{n_k}x)=d(T^lu,T^{n_k}x)$. Did I misunderstand something? – Nothingone Jun 27 at 19:10