A doubt in the proof of Uniform Boundedness Theorem as given in Kreyszig. I was looking up the proof of Uniform Boundedness Theorem. 
After having proved that the Banach space $X$ is the countable union of closed sets $\bigcup\limits_{i=1} A_i$, and hence satisfies the Baire category theorem, Kreyszig says 

Some set $A_k$ contains the open ball $B(x_0,r)\subset A_k$. 

The centre of the ball $x_0$ then goes on to play an important part in the remainder of the proof. 
Does theopen set inside $A_k$ have to be a ball? What if it is a union of balls without a specific "centre"?
Thanks in advance!
 A: The theorem of Baire states that one of the $A_k$  contains a nonempty open subset. This question shows you that every open subset is the union of open balls. If $U$ is the open set then:
$U= \bigcup_{i\in I} B(x_i,r_i) \subset A_k$
Then for every $i_0 \in I$:
$B(x_{i_0},r_{i_0}) \subset \bigcup_{i\in I} B(x_i,r_i) = U \subset A_k$
A: I would like to say something here : Generally what you say is true, open sets can be described in such a manner also (refer topology) that there can be no balls in it. But here we are dealing with a normed linear space, that is a space with a norm. 
And norm is a continuous function. $\|.\| : B \rightarrow \mathbb{R}$ which demands that open balls around origin open sets in the topology. Now translation is a homeomorphism as the vector operation is continuous again. This makes are $\epsilon$-balls in the space open.
When Normed Linear Space (Banach space is one along with completeness criterion), these sets precisely generate the whole topology. Now if you check closely, these open balls (open balls about $x$) form local base at each point $x$ in such a topology.
And then, every open set containing $x$ must contain one of them. That is precisely you have highlighted in your post.   
