x_n,$ as was already said, are bounded linear operators in $X.$ {\it The basic constant of the basis} \, $(x_n)$ is defined as the supremum of the norms $||P_N||$ over all $N;$ it is denoted by $bc\,(x_n).$ If $X$ has a basis then {\it the basic constant of the space} $X$ is not anything as the number $bc\,(X):= \inf \, bc\,(x_n),$ where {\it inf}\ is taken over all bases in the space $X.$ It is very interesting to consider this number in the case of a finite dimensional space. It is rather not hard to show that for any $n$-dimensional normed space $X$ we always have $bc\, (X)\le \sqrt n$ (the reason is that any such space is close enough to the Euclidean space $l^n_2,$ namely, is $\sqrt n$-close in the sense of "Banach--Mazur distance", but we are not going to discuss what the last phrase means, because that would lead us far enough from the main theme of our considerations, although to a very nice theory; --- it is not our aim). So, a very old question was (in connection with the last inequality) a question of the assymptotical precision of that estimate. And, by using probabilistic methods, Gluskin and Szarek showed that indeed, there is a $C>0$ so that for some sequence $(X_n)_{n=1}^\infty$ of finite dimensional spaces with $\dim X_n=n$\ one has $bc\,(X_n)\ge C\sqrt n.$ By the way, they proved that, in a sense, "no subspase of $X_n$ of large enough dimension is complemented in $X_n$" (that means, in particular, that any subspace of $X_n$ of the half dimension has the property that there is no projection from $X_n$ onto it with norm less then $c_0\sqrt n,$ for some absolute constant $c_0.)$ Well, the last mentioned finite--dimensional result leads to a very old classical question if such a situation is possible in the infinite dimensional spaces. We will give here only two striking results in this direction. Firstly, G.Pisier, answering one of the question of A.Grothendieck, showed that there exists an infinite dimensional Banach space $G$ all of whose finite dimensional subspaces lie in $G$ very bad: there is a constant $c>0$ such that for any subspace $F$ in $G$ the best projection from $G$ onto $F$ has the norm more than $c\sqrt{\dim F}$\,(! -- it is well known that every $n$-dimensional subspace of any Banach space is $\sqrt n$-complemented). Secondly, B.Maurey with a coauther showed that there exists an infinite dimensional Banach space $B$ without any complemented subspaces; more precisely, if only a subspace of $B$ is not finite dimensional itself, or is not finite codimensional, there is no linear continuous projection from $B$ onto this subspace (!!). But now, let us return from the sky filled with striking results to our bases and approximation conditions... Not knowing whether every separable Banach space has a basis, in 1955 A.Grothendieck has introduced more general notions, --- AP (approximation property) and BAP (bounded approximation property), --- to go around the basis problem and to cover the situation when the spaces under consideration are not separable (and also, in particular, to prove many facts which it was hardly to get without those additional restrictions; well, when it was possible to prove a result with only the AP, he used it, but in many cases it was needed a stronger version of the approximation in Banach spaces, so he used the MAP). Here are the definitions (in a sense, inspiring by the above discussion on bases and BAP in separable spaces, i.e. by the condition 1) and, to be exact, by conditions 2) and 3): 3) will give us the AP, while 3) with something like 2) will give the BAP and the MAP). So: They say that a Banach space $X$ has {\it the approximation property}\ (AP), if $(*)$ for each $\e>0$ and every compact set $K\sbs X$ there exists a finite rank operator $P$ in $X$ such that for each $x\in K$ one has $||Px-x||<\e.$ The space $X$ has the BAP ({\it bounded approximation property}), if there is a constant $C>0$ such that $(*)$ is fulfilled with the additional condition $||P||\le C;$ if we can take $C=1,$ then we say that the space $X$ has {\it the metric approximation property}\ (MAP). P.Enflo (we said about this a little before) proved that {\it there exists a separable reflexive space without the}\ AP, using, in particular, the following nice result of A.Grothendieck (certainly, this Grothendieck's theorem is not the main tool in the difficult proof of Enflo): {\it for reflexive Banach spaces the}\ AP {\it is just the same as the}\ MAP. This last result and some other assertions of A.Grothendieck led him to the following assumption: {\it if a weakly compact operator, acting in some Banach spaces, can be approximated, uniformly on each compact, by finite rank operators, then it also can be approximated, uniformly on each compact, by finite rank operators, whose norms are bounded by a constant, depended only on the given weakly compact operator.}\ I was successed in giving a counterexample to this assumption, constructed a separable space and a compact operator in this space, which is "approximable", but not "bounded approximable". By the way, this space has the AP, but has not the BAP. As the matter of fact, {\it the first example of the space with the}\ AP {\it and without}\ BAP {\it was found by T.Figiel and W.B.Johnson,}\ who also answered another, close to the problem, question of A.Grothendieck (related to the so called nuclearity of operators in Banach spaces). The last facts led to many interesting questions in the geometrical theory of operators in Banach spaces. But they are not in the sphere of our considerations here. We would like to mention here only the following result which generalises the Figiel-Johnson theorem. Let $I(X,X)$ be a closed subspace of the space $L(X,X)$ of all linear bounded operators in a separable space $X.$ We now know that the identity map $\id_X$ is not nessesary can be compactly approximated by a bounded sequence of finite rank operators, even if $X$ has the AP. But maybe such a situation is impossible if instead of finite rank operarors we will consider the operators from the space $I(X,X)$ (if this space $I$ is large enough, but not the whole $L(X,X)$)? However, in general, the situation here is just the same as in the Figiel-Johnson theorem, e.g.: {\it there exists a separable Banach space with the}\ AP {\it such that the identity map of this space can not be approximated, in the topology of compact convergence, by weakly compact operators.}\ With the formulation of this result (its proof is somewhat different from the one of Figiel-Johnson) I am going to finish the discussion around bases, AP, BAP etc.. And to end the talk, I would like to propose two, more that twenty years old, open questions. 1) Let $X$ be a separable Banach space. Is it true that if the dual $X^*$ has the AP, then this dual has the BAP? 2) Let $T$ be a linear bounded operator from a Banach space $X$ to a Banach space $Y$ such that there exists a continuous factorization of $T^{**}$ of the kind $$ X^{**}\to l_p \to Y^{**}, $$ where $1