Let $X$ be a seprable Hilbert space, and $\{e_n\}$ to be an orthonormal basis. Let $$\Omega=\cup_{n=1}^\infty\{(1-t)e_n+te_{n+1};0\leq t\leq 1\},$$$$T:\Omega\to R$$$$T((1-t)e_n+te_{n+1})=n+t,\ n=1,2,\cdots,\ 0\leq t\leq 1.$$ Prove that $\Omega$ is bounded, closed; and $T$ is uniformly continuous.
It is easy to verify that $\Omega$ is bounded. But I do not know how to prove the closedness. Suppose $$(1-t_k)e_{n_k}+t_ke_{n_k+1}\to x\ (k\to\infty).$$ Then up to a subsequence, $$t_k\to t_0,\ e_{n_k}\rightharpoonup \alpha,e_{n_k+1}\rightharpoonup \beta $$. What I do know is just that $$0\leq t\leq 1, (1-t)\alpha+t\beta=x.$$ How can I show that $x\in \Omega$. Also, concerning the uniform continuity of $T$, I have no idea.
The set is the union of line segments. Begin with the following observation: if $x,y\in \Omega$, then one of the following holds:
The point here is, if neither 1 or 2 holds, then when computing $x-y$, we have no cancellation: for each $n$ either $x_n$ or $y_n$ is zero. This implies 3.
Armed with the above, you should be able to prove that for every Cauchy sequence in $\Omega$, some "tail" of the sequence is contained in the union of two adjacent segments; this union is obviously closed.
For uniform continuity: try $\epsilon=\min(1,\delta)/2$. If $|x-y|<\epsilon$, you only have cases 1 and 2 above to consider.