I'm studying the following proposition but I didn't understand the proof.
If $T: X \rightarrow Y$ is an operator, then the pre-image of a closed linear subspace $B$ of $Y$ is a closed linear subspace $T^{-1}B:=\{x \in X: Tx \in B\}$ of $X$.
Proof: Let $x,\> y, \>x_n \in T^{-1}B$, that is, $Tx,\> Ty,\> Tx_n \in B$ and let $\lambda Tx \in \mathcal{F}$. Then
$$T(x + y) = Tx + Ty \in B$$ $$T(\lambda x) = \lambda Tx \in B$$ $$ x_n \rightarrow a \Rightarrow Ta = T(\lim_{n \rightarrow \infty} x_n) = \lim_{n \rightarrow \infty} Tx_n \in B $$
show that $T^{-1}B$ is a closed linear subspace.
Question: The above expressions were statements about closed linear subspace $B$ of $Y$ -- such that continuous linear transformation $T$ preserve vector addition, scalar multiplication and convergence. I don't get it how does it show that $T^{-1}B$ is a closed linear subspace.
Thank you!
Since $T(x+y)\in B$, $x+y\in T^{-1}B$. Similarly, the other statements show that $\lambda x\in T^{-1}B$ and $a\in T^{-1}B$. Together these show that $T^{-1}B$ is a closed subspace of $X$.