The Banach-Mazur distance is not reached Let $X,Y$ be isomorphic Banach spaces.
The Banach-Mazur distance: 
$$
d(X,Y)=\inf\{\|T\| \cdot \|T^{-1}\|: T:X\longrightarrow Y \ \text{is an isomorphism} \}$$
can be rewritten as:
$$
d(X,Y)=\inf\{\|T^{-1}\|: T:X\longrightarrow Y \ \text{is an isomorphism}, \|T\|=1\}
$$
If $X,Y$ are finite dimensional spaces the infimum is reached. 
But if $X,Y$ are infinite dimensional spaces the infimum is reached ?
Any hints would be appreciated.
 A: Let's follow the hint given by Wojtaszczyk.

Take $(p_n)$ and $(q_n)$ two disjoint sequences, dense in $[1,1.5]$ such that $p_1=1$. Take $X=\left(\sum \ell_{p_n}^5\right)_2$ and $Y =\left(\sum \ell_{q_n}^5\right)_2$.

The Banach-Mazur distance is $1$ because for every $\epsilon>0$ the interval $[1,1.5]$ can be partitioned into subintervals of size $\epsilon$, each of which meets both sequences countably many times. 
Now, it seems to me that the reason $Y$ does not contain an isometric copy of $\ell_1^5$ is that $Y$ is strictly convex. (If I'm right then neither $1.5$ nor $5$ are of importance; $1.5$ could be any  number in $(1,\infty)$ and $5$ could be an integer $\ge 2$.) 
Indeed, suppose that $(y_n)$  and $(z_n)$ are  two nonzero elements of $Y$ such that $\|(y_n+z_n)\|_Y = \|(y_n)\|_Y+\|(z_n)\|_Y$, meaning that
$$\sqrt{\sum \left\| y_n+z_n \right\|_{q_n}^2} =  \sqrt{\sum \left\|y_n\right\|_{q_n}^2} +  \sqrt{\sum \left\|z_n\right\|_{q_n}^2} \tag1$$
By the triangle inequality and Minkowski inequality for $\ell^2$,
$$\sqrt{\sum \left\| y_n+z_n \right\|_{q_n}^2}
\le \sqrt{\sum (\left\| y_n\|_{q^n}+ \|z_n \right\|_{q_n})^2}\le \sqrt{\sum \left\|y_n\right\|_{q_n}^2} +  \sqrt{\sum \left\|z_n\right\|_{q_n}^2} \tag{2}$$
where equality must hold throughout by (1). 
Since $\ell_2$ is strictly convex, equality in the second half of (2) implies that there exists $\lambda> 0$ such that  $ \|z_n\|_{q_n}=\lambda \|y_n\|_{q_n}$ for all $n$. Since each $\ell_{q_n}$ is also strictly convex, equality in the first half of (2) implies that $z_n=\lambda y_n$. Thus, $Y$ is strictly convex.
A: For completeness, I add the details of the proof of $d_{BM}(X,Y)=1$ for the example suggested by Wojtaszczyk. The key is the following
Lemma. The identity map $\ell_p^n\to \ell_q^n $ has norm $1$ when $p\le q$, and norm $n^{1/p-1/q}$ when $p\ge q$. 
Proof. (a) Suppose $p\le q$. Let $x$ be a vector of unit $p$-norm. Since each coordinate of $x$ has magnitude at most $1$, it follows that $|x_i|^q\le |x_i|^p$. Hence $\|x\|_q\le 1$.
(b) Suppose  $p\ge q$. By Jensen's inequality, applied to the convex function $\phi(t)=t^{p/q}$, implies
$$
\frac{1}{n} \sum (|x_i|^q)^{p/q} \ge \left(\frac{1}{n} \sum |x_i|^q\right)^{p/q}  
$$
hence $\|x\|_q \le n^{1/q-1/p}\|x\|_p$.

In the construction considered here, $n=5$ and $p,q\in [1,1.5]$. Divide $[2/3,1]$ into subintervals of length $<\epsilon$, for small $\epsilon$. If $1/p$ and $1/q$ fall into the same subinterval, the Banach-Mazur distance between $\ell_p^5$ and $\ell_q^5$ is at most $5^\epsilon=1+O(\epsilon)$.
Since the sequences $p_k$ and $q_k$ are dense in $[1,1.5]$, each of subintervals mentioned above contains countably many exponents of either kind. Use a bijection between them to define an isomorphism. The norm of this isomorphism is at most $5^\epsilon$ in either direction.
