Extension of bounded linear operators Let $X,Y$ be Banach spaces, and let $X_0$ be a subspace of $X$ (by subspace I mean a closed linear set). Consider the set $Ext(X_0,Y)$ of all bounded linear operators $A_0:X_0\to Y$ which have an extension to a bounded linear operator $A:X\to Y$ 
(not necessarily with the same norm).
Question: is $Ext(X_0,Y)$ closed in $B(X_0,Y)$ (with the usual operator norm)?
Equivalently, is the set of all bounded linear operators $A_0:X_0\to Y$ which have no extension to a bounded linear operator $A:X\to Y$ open in $B(X_0,Y)$?
If $Ext(X_0,Y)$ is not closed (in general), for which $X,X_0;Y$ this set is closed in $B(X_0,Y)$?
I will be very grateful for any remarks and comments.
 A: The set $Ext(X_0, Y)$ will be rarely closed, I believe. (Perhaps you want to use a different symbol for this set, as $Ext$ has its own meaning.)
Suppose that $X$ is a Banach space which is complemented in $X^{**}$, $X$ is isomorphic to $X\oplus X$, $X$ has a Schauder basis and it contains an uncomplemented copy of itself, $X_0$ say. Then there is a compact operator $K\colon X_0\to X$ which does not extend to $X$. (This is implicit in in the proof of Lemma 5.7 here; apologies for self-advertisement). On the other hand, $X_0$ has the approximation property, so $K$ can be approximated by finite-ranks which are extendable. 
The classical spaces $\ell_p$ and $L_p$ for $p\in [1,2)\cup (2,\infty)$ satisfy the above assumptions. The assumption that $X\cong X\oplus X$ can be dropped but things get messier then.
Note that the above situation with compact operators cannot happen if $Y$ is a $\mathscr{L}_\infty$-space as in this case we have the Grothendieck–Lindenstrauss theorem about extensions of compact operators.
A: Tomek answered your question, but here are some further comments. $Ext(X_0,Y)$ is the image of $L(X,Y)$ under the restriction map, the kernel of which is a closed subspace, so $Ext(X_0,Y)$ is complete under a stronger norm. Thus the only way $Ext(X_0,Y)$ can be complete in the operator norm is for the two norms to be equivalent. (Otherwise, as Tomek pointed out, there will be a non extendable compact operator.)  Now every finite rank operator from $X_0$ to $Y$ is extendable, so for $Ext(X_0,Y)$ to be complete in the operator norm, there must be a constant $C$ s.t. every finite rank norm one operator from $X_0$ to $Y$ has an extension to an operator of norm at most $C$.  This condition is also sufficient if $Y$ is reflexive (or, more generally, is complemented in $Y^{**}$).  Indeed, direct the finite dimensional subspaces of $X_0$ by inclusion.  Given $T:X_0 \to Y$ of norm one, consider for $E\subset X_0$ finite dimensional an extension $T_E$ of the restriction of $T$ to $E$ with $\|T_E\|\le C$.  This net has a subnet that converges in the weak$^*$ operator topology to an extension of $T$ to an operator from $X$ to $Y^{**}$ that has norm at most $C$. 
