**EDIT:** In my original post, I showed that $\mathrm{cov}(\mathcal N) > \mathfrak{x}_{lac}$ in the random model. Upon further reflection, I think we can prove a stronger result, with an arguably easier (but completely different) proof:

*Theorem:* $\mathfrak{x}_{lac} \leq \mathrm{non}(\mathcal N)$.

Note that this implies $\mathfrak{x}_{lac} < \mathrm{cov}(\mathcal N)$ in the random model, and it also implies the consistency of $\mathfrak{x}_{lac} < \mathfrak{d}$.

In addition to this theorem, let me also point out that it is consistent to have $\mathfrak{a} < \mathrm{cov}(\mathcal M)$. (See Corollary 2.6 in this paper of Brendle.) Therefore the lower bound $\mathrm{cov}(\mathcal M) \leq \mathfrak{x}_{lac}$ mentioned in the post already implies $\mathfrak{a}$ is not an upper bound for $\mathfrak{x}_{lac}$.

*Proof of the theorem:* Suppose we form an infinite set $B$ by choosing from each interval of the form $[2^k,2^{k+1})$ exactly one integer $b_k$ at random, and then taking $B = \{b_k :\, k \in \omega \}$. (By "at random" I mean that we choose with the uniform distribution, so each integer in $[2^k,2^{k+1})$ has probability $1/2^k$ of being selected.) I claim that if $A$ is lacunary, then it is almost surely true that $A \cap B$ is finite.

To see this, fix some $c > 1$ and some $n_0 \in \omega$ such that if $a$ and $b$ are consecutive members of $A$ above $n_0$, then $b/a > c$. If $k$ is large enough that $n_0 < 2^k$, then this implies there are at most $log_c(2)$ members of $[2^k,2^{k+1})$ in $A$. This implies that the probability of choosing $b_k \in A$ is $\log_c(2)/2^k$ when $k$ is large enough. It follows that the probability of there being $>K$ members of $A$ in $B$ (when $K$ is large) is $\sum_{k > K} \log_c(2)/2^k = \log_c(2)/2^K$. Since this goes to $0$ as $K$ goes to infinity, the probability of $A \cap B$ being infinite is $0$.

The idea of choosing $B$ randomly, one point at a time, like this can be formalized by defining a probability measure on a Polish space, where points of the space correspond to possible choices of the sequence of $b_k$'s. What the previous paragraph shows is that in this probability space, the set of all $B$'s with $A \cap B$ infinite form a null set, for any given lacunary set $A$. Hence any non-null subset of this probability space will contain a $B$ with $A \cap B$ finite. Since this holds for any $A$, we see that any non-null subset $X$ of this probability space contains, for *any* lacunary set $A$, some infinite $B$ with $A \cap B$ finite. The smallest possible size of such a set $X$ is $\mathrm{non}(\mathcal N)$.

*QED*

One more observation: The strict inequality $\mathfrak{x}_{lac} < \mathrm{non}(\mathcal N)$ is also consistent, so this upper bound cannot be improved to an equality. To see this, begin with a model of Martin's Axiom $+ \, \neg \mathsf{CH}$, and then do a legnth-$\omega_1$, finite support iteration of the eventually different reals forcing. It is not difficult to see that this forcing will make $\mathfrak{x}_{lac} = \aleph_1$ in the extension. But the iteration is $\sigma$-centered, and forcing with a $\sigma$-centered poset over a model of $\mathsf{MA}$ does not change the value of $\mathrm{non}(\mathcal N)$. Thus we get $\mathfrak{x}_{lac} < \mathrm{non}(\mathcal N)$ in the extension.

**Original post:**

It is not provable that $\mathrm{cov}(\mathcal N) \leq \mathfrak{x}_{lac}$, because $\mathrm{cov}(\mathcal N) > \mathfrak{x}_{lac}$ in the random model.

To see this, let me sketch an argument that after forcing to add any number of random reals (in the usual way, via a measure algebra), the collection $[\omega]^\omega \cap V$ of infinite subsets of $\omega$ in the ground model has the property described in the definition of $\mathfrak{x}_{lac}$. That is: for every lacunary set $A \subseteq \omega$ in the extension, there is an infinite $B \subseteq \omega$ in the ground model such that $A \cap B$ is finite.

We work in the ground model. Suppose $\dot A$ is a name for an infinite lacunary set in the extension. There is some fixed $c > 1$ and $n_0 \in \omega$, and some forcing condition $p$, such that $p \Vdash$ if $a$ and $b$ are consecutive elements of $\dot A$ above $n_0$, then $b/a > c$.

I claim that for every $\varepsilon > 0$, there are arbitrarily large $n \in \omega$ such that $m(p \wedge [n \in \dot A]) < \varepsilon$.

(Note: Here I'm using the standard notation for forcing with measure algebras. If $\varphi$ is any well-formed formula in the forcing language, then we write $[\varphi]$ to mean the supremum of all the conditions forcing $\varphi$, and $m([\varphi])$ denotes the measure of this supremum. Roughly, you may think of $m([\varphi])$ as the probability that $\varphi$ ends up being true in the forcing extension.)

To prove my claim, suppose, aiming for a contradiction, that it is false. Then there is some $\varepsilon > 0$ and some $N \in \omega$ such that $m([n \in \dot A]) \geq \varepsilon$ for all $n \geq N$. But this is just another way of saying that the "expected size" of $A \cap \{n\}$ is $\geq\varepsilon$ for all $n \geq N$. By the linearity of expectation, this means the expected size of $A \cap \{k,k+1,\dots,\ell-2,\ell-1\}$ is $\geq (\ell-k)\varepsilon$ whenever $\ell > k > N$. But by our choice of $p$, if $k \geq N$ and $\ell < ck$, then $p \Vdash |A \cap \{k,k+1,\dots,\ell-2,\ell-1\}| \leq 1$. Since $c > 1$, this yields a contradiction for sufficiently large $k$, namely $k > N,2/c\varepsilon$.

Using this claim, we can now find an infinite ground model set almost disjoint from the set named by $\dot A$ in the extension. Using the claim, we may find for each $k \in \omega$ some $n_k > n_{k-1}$ such that $m([n_k \in \dot A]) < m(p)/2^{k+2}$. Now let $p' = p - \bigvee_{k \in \omega}[n_k \in \dot A]$. Then $m(p') \geq m(p) - \sum_{k \in \omega}m([n_k \in \dot A]) > m(p)/2 > 0$, so $p'$ is a condition in our measure algebra, and $p'$ forces $\dot A$ to be disjoint from $\{n_k :\, k \in \omega \}$ (because it extends the complement of each $[n_k \in \dot A]$).

This shows that it is impossible to have a name $\dot A$ for a lacunary set such that $\dot A$ is forced to have infinite intersection with every infinite subset of $\omega$ from the ground model. Therefore there is no such set.