A question about a dense subset in Banach space. Let $X,Y$ be Banach spaces, $T:X\to Y$,unbounded linear operator.
How to prove that there is a natural number $n$,the set $\{x:\|Tx\|\le n\|x\|\}$ is dense in $X$?
 A: The claim is true (see the proof below using Baire's Category Theorem). But note that it is (without the axiom of choice) hard to come up with an unbounded operator $T$ that is defined on the whole Banach space $X$.
For $n\in\mathbb{N}$ define
$$
A_{n}:=\left\{ x\in X \mid \left\Vert Tx\right\Vert \leq n\cdot\left\Vert x\right\Vert \right\} .
$$
Note that $0\in A_{n}$ for all $n\in\mathbb{N}$ and that we have
$rA_{n}\subset A_{n}$ for all $r>0$ and $n\in\mathbb{N}$.
Also note that we have
$$
X=\bigcup_{n\in\mathbb{N}}A_{n}\subset\bigcup_{n\in\mathbb{N}}\overline{A_{n}}.
$$
The Baire Category Theorems shows that there is $n_{0}\in\mathbb{N}$,
$x_{0}\in X$ and $\varepsilon>0$ such that
$$
B_{\varepsilon}\left(x_{0}\right)\subset\overline{A_{n_{0}}}
$$
holds.
Choose $m\in\mathbb{N}$ with
$$
\frac{2}{\varepsilon}\cdot\left[n_{0}\cdot\left(\left\Vert x_{0}\right\Vert +\varepsilon\right)+\left\Vert Tx_{0}\right\Vert \right]\leq m.
$$
Now let $x\in X$ be arbitrary with $\frac{\varepsilon}{2}<\left\Vert x\right\Vert <\varepsilon$,
so that $x_{0}+x\in B_{\varepsilon}\left(x_{0}\right)\subset\overline{A_{n_{0}}}$
holds. Thus, there is a sequence $\left(y_{n}\right)_{n\in\mathbb{N}}$
in $A_{n_{0}}$ such that $y_{n}\rightarrow x_{0}+x$ holds. This
yields $y_{n}-x_{0}\rightarrow x$ and
$$
\begin{eqnarray*}
\left\Vert T\left(y_{n}-x_{0}\right)\right\Vert  & \leq & \left\Vert Ty_{n}\right\Vert +\left\Vert Tx_{0}\right\Vert \\
 & \leq & n_{0}\cdot\left\Vert y_{n}\right\Vert +\left\Vert Tx_{0}\right\Vert \\
 & = & \underbrace{\frac{n_{0}\cdot\left\Vert y_{n}\right\Vert +\left\Vert Tx_{0}\right\Vert }{\left\Vert y_{n}-x_{0}\right\Vert }}_{=:\alpha_{n}}\cdot\left\Vert y_{n}-x_{0}\right\Vert .\qquad\left(\ast\right)
\end{eqnarray*}
$$
Now note that $\frac{\varepsilon}{2}<\left\Vert x\right\Vert <\varepsilon$
yields
$$
\begin{eqnarray*}
\alpha_{n}\xrightarrow[n\rightarrow\infty]{}\frac{n_{0}\cdot\left\Vert x_{0}+x\right\Vert +\left\Vert Tx_{0}\right\Vert }{\left\Vert x\right\Vert } & < & \frac{2}{\varepsilon}\cdot\left[n_{0}\cdot\left\Vert x_{0}+x\right\Vert +\left\Vert Tx_{0}\right\Vert \right]\\
 & \leq & \frac{2}{\varepsilon}\cdot\left[n_{0}\cdot\left(\left\Vert x_{0}\right\Vert +\left\Vert x\right\Vert \right)+\left\Vert Tx_{0}\right\Vert \right]\\
 & < & \frac{2}{\varepsilon}\cdot\left[n_{0}\cdot\left(\left\Vert x_{0}\right\Vert +\varepsilon\right)+\left\Vert Tx_{0}\right\Vert \right]\\
 & \leq & m.
\end{eqnarray*}
$$
We thus get $\alpha_{n}<m$ for $n$ large enough. The estimate marked
with $\left(\ast\right)$ above thus shows $y_{n}-x_{0}\in A_{m}$
for $n$ large enough and thus $x=\lim_{n}\left(y_{n}-x_{0}\right)\in\overline{A_{m}}$.
We have thus established
$$
B_{\varepsilon}\left(0\right)\setminus\overline{B_{\varepsilon/2}}\left(0\right)\subset\overline{A_{m}}.
$$
This implies
$$
B_{r\varepsilon}\left(0\right)\setminus\overline{B_{r\varepsilon/2}}\left(0\right)\subset r\overline{A_{m}}=\overline{rA_{m}}\subset\overline{A_{m}}
$$
for all $r>0$. But 
$$
X\setminus\left\{ 0\right\} =\bigcup_{r>0}\left[ B_{r\varepsilon}\left(0\right)\setminus\overline{B_{r\varepsilon/2}}\left(0\right)\right],
$$
so that $X\subset\overline{A_{m}}$ follows. This completes the proof.
