# Confusion about equivalence of continuity and boundedness of linear operators.

I am confused with a part of the proof for the following theorem.

Theorem : Let $$E$$ and $$F$$ be normed spaces. Let $$T: E\rightarrow F$$ be linear. Then $$T$$ is continuous if it is bounded.

Let $$T$$ be continuous, if it is bounded we are done so suppose it is unbounded. I am confused with the following statement:

Since $$T$$ is unbounded for every $$n$$ we can find and $$x_n$$ such that $$\|Tx_n\|>n$$ and $$\|x_n\|=1$$.

Part of the proof that I am confused with:

Since $$T$$ is unbounded it is clearly through that we can find arbitrarily large $$n$$ with $$\|Tx_n\|>n$$, but I do not see how we can simultaneously require $$\|x_n\|=1$$.

Additional notes on my confusing that may help when explaining this to me: I believe I am missing a common idea or trick used in functional analysis as I am also confused when trying to show that the following are equal $$\sup_{x\in E,\|x\|\leq 1}|f(x)|=\sup_{x\in E,\|x\|= 1}|f(x)|$$

You are dealing with linear operators between normed spaces $$X$$ and $$Y$$ here. And we say that such an operator $$T$$ is bounded if there is some $$M\geqslant0$$ such that$$(\forall x\in X):\bigl\lVert T(x)\bigr\rVert\leqslant M\lVert x\rVert.$$In particular, $$\lVert x\rVert=1\implies\bigl\lVert T(x)\bigr\rVert\leqslant M$$.