Problem in proof of open mapping theorem? I was doing proof of open mapping theorem from the book Walter Rudin real and complex analysis book and struck at one point. Given if $X$ and $Y$ are Banach spaces and $T$ is a bounded linear operator between them which is $\textbf{onto}$. Then to prove $$T(U) \supset \delta V$$ where $U$ is open unit ball in $X$ and $\delta V = \{ y \in Y : \|y\| < \delta\}$.
Proof- For any $y \in Y$ since map is onto, there exist an $x \in X$ such that $Tx = y$. It is also clear that if $\|x\| < k$, then $y \in T(kU)$ for any $k$. Clearly $$Y = \underset{k \in \mathbb{N}}{\cup}  T(kU) $$ But as $Y$ is complete, by Baire category theorem it can't be written as countable union of nowhere dense sets. So there exist atleast one $k$ such that $ T(kU)$ is not nowhere dense.  Thus this means $$(\overline{T(kU)})^0 \ne \emptyset$$ i.e. $ T(kU)$ closure has non empty interior. Let $W$ be open set contained in closure of  $T(kU)$. Now for any $w \in W \implies w \in \overline{T(kU})$, so every point of $W$ is the limit of the sequence $\{Tx_i\}$, where $x_i \in kU$, Let us now fix $W$ and $k$.
Now choose $y_0 \in W$ and choose $\eta > 0$, so that $y_0+y \in W$ if $\|y\| < \eta$. This can be done as $W$ is open set, so every point of it has some neighborhood also there. Now as $y_0 , y_0+y \in W$ from above paragraph  there exist sequences $\{x_i'\}$ and $\{x_i''\}$ in $kU$ such that $$T(x_i') \to y_0 \qquad T(x_i'') \to y_0+y \quad as \ i \to \infty$$ Set $x_i = x_i'-x_i''$. Then clearly $$\|x_i\| \leq \|x_i'\| + \|x_i''\| < 2k$$ and $T(x_i) \to y$. Since this holds for every $y$ with $\|y\|< \eta$. 
Now it is written that, the linearity of $T$ shows that following is true for $\delta = \dfrac{\eta}{2k}$ 
To each $y \in Y$ and to each $\epsilon > 0$ there corresponds an $x \in X$ such that $$\|x\| \leq \delta^{-1}\|y\| \quad \text{and} \quad \|Tx-y\| < \epsilon \quad (1)$$ How does this follows?
This proof is given in Walter rudin 3rd edition on page 112
 A: Notice that if $||y|| < \eta$ then $y \in \eta V$, and the last thing to be proved in your question was that for all $y \in \eta V$ there exists a sequence of $x_n \in 2kU$ (which implies that $Tx_n \in T(2kU)$) such that $Tx_n \to y$, i.e. there is a sequence in $T(2kU)$ that converges to the limit point $y$, so that $y$ is in the closure $\text{cl}[{T(2kU)}]$. This means that $\eta V \subseteq \text{cl}[{T(2kU)}]$. If $y \in \eta V$ then for all $\epsilon >0$ there exists an $N$ such that for all $n \ge N$, $||Tx_n-y||<\epsilon$. Fix any $r >0$ and notice that $||T(rx_n)-ry|=||rTx_n-ry||=r||Tx_n-y||<r\epsilon$. This means that for all $ry \in r\eta V$ there exists a sequence $Trx_n \in T(2rkU)$ which converges to $ry$, hence $r\eta V \subseteq \text {cl} [T(2rkU)]$.
 
Set $r = \frac {t}{\eta}$ and you get $$tV \subseteq \text {cl} [T(\delta^{-1}tU)]$$
Since every $y \in Y$ is contained in $||y||V\subseteq \text {cl} [T(\delta^{-1}||y||U)]$ there exists a sequence such that $||x_n|| \le \delta ^{-1}||y||$ and $||Tx_n-y||  < \epsilon$. 
A: For any $y\in Y$, set $y^\prime = \dfrac {\eta y} {\|y\|}$.
$y_0 +y^\prime \in \overline W$, so we can find $x^\prime\in 2kU$ such that $\|Tx^\prime - y^\prime\| < \dfrac {\eta\epsilon}{\|y\|}$.
Then (1) holds for $y$, with $x = \dfrac {\|y\| x^\prime}\eta$.  
Or, just choose $\eta$ such that $y_0+y\in W$ if  $\|y\| \leq \eta$ instead.  
