Let $(H_n)^{\infty}_{n=1}$ be a family of Hilbert spaces and let $H=\oplus_{n=1}^{\infty}H_n$ be the Hilbert space which is the direct sum of the family $(H_n)^{\infty}_{n=1}$. If $T_n \in L(H_n)$ and $\sup_n \|T_n\|< \infty$ for all $n\geq 1$, define $T\in L(H)$ by $$T(x_1, x_2,...,x_n,...)=(T_1x_1, T_2x_2,...,T_nx_n,...).$$ Given this setting, I am stuck with the following question.
Show that $\overline{\cup^{\infty}_{n=1} \sigma(T_n)} \subset \sigma(T).$
Any ideas would be appreciated!
Assume $\lambda\not\in \sigma(T)$, then $T-\lambda I$ is a bounded continuous operator. Note that it is a purely algebraic fact that an operator $A:H\to H$ is invertible if and only if $\pi_n\circ A\circ \iota_n:H_n\to H_n$ is invertible, where $\pi_n H\to H_n$ is the canonical projection and $\iota_n:H_n\to H$ is the natural inclusion. Moreover, in this case \begin{align*} A^{-1}(x_1,a_2,\dots) &= ((\pi_1\circ A\circ \iota_1)^{-1}x_1,(\pi_2\circ A\circ \iota_2)^{-1}x_2,\dots)\\ &= ((\pi_1\circ A^{-1}\circ \iota_1)x_1,(\pi_2\circ A^{-1}\circ \iota_2)x_2,\dots). \end{align*} Also note that in this case, $T_n-\lambda I_n=\pi_n\circ (T-\lambda I)\circ\iota_n$, where $I_n:H_n\to H_n$ is the identity operator. The open mapping theorem implies that $T_n-\lambda I_n:H_n\to H_n$ has a bounded inverse, thus $\lambda\not\in \sigma(T_n)$ for all $n$. Thus we have proved that $$ \mathbb{C}\setminus \sigma(T) \subseteq \bigcap_{n\geq 1} \mathbb{C}\setminus\sigma(T_n) $$ and thus $$ \bigcup_{n\geq 1} \sigma(T_n)\subseteq \sigma(T). $$ But it is easy to see that $T$ is bounded, and thus $\sigma(T)$ is a compact subset of $\mathbb{C}$, in particular it is closed, and then $$ \overline{\bigcup_{n\geq 1} \sigma(T_n)}\subseteq \sigma(T), $$ as desired.