Question on the proof of Open mapping Theorem

Question

I am studying on Open Mapping Theorem. I am stuck at a point in the proof.

Open Unit Ball: A bounded linear operator T from a Banach space X onto a Banach space Y has the property that the image $T(B_0)$ of the open unit ball $B_0 = B(0;1) \subset X$ contains an open ball about $0 \in Y$.

Proof

Consider the open ball $B_1 = B(0; 0.5) \subset X $. Any fixed $ x \in X $ is in $kB_1$ with sufficienly large k. Hence, $ X = \bigcup_{k=1}^{\infty} kB_1$, ...

I can know $X \subset \bigcup_{k=1}^{\infty} kB_1$, however, I do not understand how $ X = \bigcup_{k=1}^{\infty} kB_1$ is possible.

Answer 1

To show $X = \bigcup_{k=1}^{\infty} kB_1$ we must check both $\subset$ and $\supset$ relations between these.

$\subset$ : for every $x\in X$ there is $k$ such that $x\in kB_1$. E.g., $k$ could be any integer greater than $\|x\|$.

$\supset$ : the Banach space $X$ is our Universe here; no elements from outside of it enter the proof. The unit ball $B_1=\{x\in X : \|x\|<1 \}$ is a subset of $X$. So is $kB_1$, since linear spaces are closed under scalar multiplication.