Recall that the $i$*-weight* of a Tychonoff space $X$, $iw(X)$, denotes the minimal weight of all Tychonoff spaces onto which $X$ can be condensed. A standard fact about this cardinal number is that the relations $\text{log}(X)\leq iw(X) \leq nw(X)$ always hold. Furthermore, if $X$ is an infinite discrete space, then $\text{log}(X)$ and $iw(X)$ are actually the same.


Is it true that all infinite metrizable spaces $X$ satisfiy the equality $\text{log}(X)=iw(X)$?



$i$-weight of a metrizable space
Countable powers of copies of $H(\kappa)$, the hedghog space of spininess $\kappa$, are universal for metrizable spaces of weight $\kappa$ (i.e., any metrizable space of weight $\kappa$ embeds into such a countable power as a subspace). Note first that $H(2^\kappa)$ condenses onto a subspace of $[0,1]^\kappa$ (it is easier to see geometrically how $H(2^\omega)$ condenses onto the unit disc using the usual geometric picture of the hedgehog, but, in general, to map $H(2^\kappa)$ onto $[0,1]^\kappa$: for each $f\in 2^\kappa\setminus \{0\}$, let $I\_f=\{t\cdot f:0\leq t\leq 1\}$ and map the spine $I\_f$ in $H(2^\kappa)$ to the subspace $I\_f$ in the product space $[0,1]^\kappa$). The weight of this subspace is $\leq\kappa$. Now, if a metrizable space has cardinality $2^\kappa$, it embeds into $H(2^\kappa)^\omega$ which in turn condenses onto a subspace of $(I^{\kappa})^\omega$ and so $iw(X)\leq \kappa=Log(X)$ and by the reverse inequality mentioned in your question you get the equality you were asking for.


