Show that $E$ is closed. Let $X$ be a normed linear space .Let $T_n$ be  a sequence of continuous linear operators on $X$ such that $\sup_n \|T_n\|<\infty$. Let $E=\{x:T_n x $ is Cauchy$\}$. Show  that $E$ is closed.
My try:
Let $x_n$ be a sequence in $E$ such that $x_n\to x\implies T_n x_n\to T_n x\forall n\in \Bbb N$ (Since $T_n$ is continuous ) $\rightarrow (1)$.
Now $\|T_n x-T_m x\|\leq \|T_n x- T_n x_n \|+\|T_n x_n- T_m x_n\|+\|T_m x_n-T_m x\|$.
Again $\|T_n x- T_n x_n \|\to 0$ as $ n\to \infty$ by $(1)$.
Similarly $\|T_m x_n-T_m x\|\to 0$ as $ n\to \infty$ by $(1)$
Again $T_nx_n$ is Cauchy for all $n$. So $\|T_nx_n-T_m x_n\|\to 0$ as $n,m\to \infty$
So $\|T_n x-T_m x\|\to 0$ as $n,m\to \infty\implies T_n x $ is Cauchy.
Is the above proof correct? But the problem is I haven't used that $\sup_n \|T_n\|<\infty$. Please suggest edits if required
 A: The proof is wrong, but it isn't easy to see why: Let us use your proof formulating in terms of $\epsilon$-definition:
Given $\epsilon>0$, there is an $N_n$ (a natural number depending on $n$) such that if $r\ge N_n$ then $\|T_nx-T_nx_r\|<\frac{\epsilon}{3}$
Given $\epsilon>0$, there is an $N_m$ (a natural number depending on $m$) such that if $r\ge N_m$ then $\|T_mx-T_mx_r\|<\frac{\epsilon}{3}$
Given $\epsilon>0$, there is an $N_r$ (a natural number depending on $r$) such that if $n,m\ge N_r$ then $\|T_nx_r-T_mx_r\|<\frac{\epsilon}{3}$
But now, naturally you want to choose $U=\max\{N_n,N_m,N_r\}$ and take $n,m,r \ge U$. But $U$ is undefined!! (cause $U$ depends on $n,m,r$ and $n,m,r$ depend again on a triple $n,m,r$). Then you can't choose $U$ such that the above requirements are satisfied.
But, using the hypothesis $\|T_n\|\le M$, we have
$$\|T_nx-T_mx\|\le\|T_nx-T_nx_r\|+\|T_mx-T_mx_r\|+\|T_nx_r-T_mx_r\|\\\le\|T_n\|\|x-x_r\|+\|T_m\|\|x-x_r\|+\|T_nx_r-T_mx_r\|\\\le 2M\|x-x_r\|+\|T_nx_r-T_mx_r\|$$
Now, for $\epsilon>0$, since $x_r\to x$, there is $N$ only depending on $\epsilon$ such that $\|x-x_r\|<\frac{\epsilon}{2\times 2M}$ for $r\ge N$. Fix $r=N$ (note that this step is not plausible in your argument), and since $\{T_nx_N\}$ is Cauchy, there is an $M$ depending only in $N$ such that $\|T_nx_N-T_mx_N\|<\frac{\epsilon}{2}$ for $n,m\ge M$.
Thus, if $n,m\ge M$ we have:
$$\|T_nx-T_mx\|\le 2M\|x-x_N\|+\|T_nx_N-T_mx_N\|\\\le\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$.
Done.
Note that the idea was "take all values depending only on $r$ and not on $n,m$".
