# Open mapping lemma - are these versions equivalent?

Here is a version the Open Mapping Lemma given in class :

Let $X$ be a Banach space and $Y$ be a normed space. Let $T : X\rightarrow Y$ be a bounded linear map. Assume there exist $M \geq 0$ and $0 \leq \delta < 1$ such that $T(MB_X)$ is $\delta$-dense in $B_Y$. Then $T$ is surjective. More precisely for all $y \in Y$ there exists $x \in X$ such that $y = T(x)$ and $\|x\| \leq \frac{M}{1-\delta}\|y\|$, i.e. $T\left(\frac{M}{1-\delta}B_X\right) \supseteq B_Y$. Moreover $Y$ is complete.

Note that $A$ is said to be $\delta$-dense in $B$ if $\forall b \in B$, $\exists a \in A$ such that $d(a,b) \leq \delta$.

I am wondering if this is stronger/weaker/equivalent/independent than/to/from the following Open Mapping Theorem (strong version) :

Let $T$ be a bounded linear map from a Banach space $X$ into a normed space $Y$. If the image $T(X)$ is nonmeager in $Y$ then $T$ is surjective and an open mapping. Moreover $Y$ is complete.

Any hint ?

For a subsets of topological vector space $X$ the following theorem hols true.
Theorem: If $A,B,C\subset X$ and $B$ is bounded $C$ is closed and convex and $A+B\subset C+B,$ then $A\subset C$ e have $B_Y \subset T(MB_X )+ \delta B_Y ,$ hence $$\delta B_Y +(1-\delta ) B_Y \subset \overline{T(MB_X )}+ \delta B_Y ,$$ and from the theorem we obtain that $$(1-\delta ) B_Y \subset \overline{T(MB_X )} ,$$ hence the image $T(X)$ is not meager in $Y.$