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 $d(Tx,Ty)<d(x,y)$ 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 $$d(u,T^{m}x)=d(T^lu,T^{n_k+l}x)<d(u,T^{n_k}x)<\varepsilon.$$ 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 $T$.

  • $\begingroup$ Why is $d(u,T^mx)=d(T^lu, T^{n_k+l}x)?$ $\endgroup$ – Adrian Keister Jun 27 at 18:43
  • $\begingroup$ 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? $\endgroup$ – Nothingone Jun 27 at 19:10

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