Can we cover a self similar set using another self similar set with smaller dimension? Suppose we have two self similar sets $A$ and $B$ constructed as follows:

Let $\varphi_i(x):=\frac{x+i}{5},i=0,1,2,3,4$ be similarities, then $A$ and $B$ are self similar sets with:

$$A=\varphi_0(A)\cup\varphi_2(A)\cup\varphi_4(A)\text{ and }B=\varphi_1(A)\cup\varphi_3(A).$$
Can we cover $A$ with countably many translates of the set $B$?

I believe not, but I am stuck on one small implication.

My attempt:

Using the open set condition on $(0,1)$ we can easily calculate the dimension of $A$ and $B$ using Morans Theorem and show that $\dim_H(B)&lt;\dim_H(A).$
Since a translation is a bi-Lipshitz map, it preserves the Hausdorff dimension (informally, moving $B$ doesn't change its Hausdorff dimension). Also, the dimension of a countable union of sets is the supremum of the dimension of sets, hence any countable union of translations of $B$ have dimension $\dim_H(B)$ which is strictly less than $\dim_H(A)$. 

So finally, can I conclude that if the dimension of countably many translations of $B$ is still less than the dimension of $A$, we can never cover it this way?
The Hausdorff dimension has the following relevant properties:


Countable stability: Let $\{ E_j \}_{j\in \mathbb{N}}$ be a countable collection of sets.  Then
$$ \dim_H\left( \bigcup_{j\in\mathbb{N}} E_j \right) \le \sup_{j\in\mathbb{N}} \dim_{H}(E_j).$$
Monotonicity: If $E \subseteq F$, then $\dim_H(E) \subseteq \dim_H(F)$.


Suppose that $A$ and $B$ are arbitrary subsets of some Euclidean space $\mathbb{R}^n$ such that
$$ \dim_{H}(B) &lt; \dim_H(A). $$
Assume for contradiction that there exists a countable collection of translations $\{\tau_j\}_{j\in\mathbb{N}}$ such that
$$ A \subseteq \bigcup_{j\in\mathbb{N}} \tau_j B.$$
That is, assume that $A$ may be covered by a countable number of copies of $B$.  Translation is an isometry, hence it preserves the Hausdorff dimension.  Thus $\dim_{H}( \tau_j B) = \dim_H(B)$ for all $j$.  The Haudsorff dimension is countably stable and monotonic, thus
$$ \dim_{H}(A)
\le \dim_{H}\left( \bigcup_{j\in\mathbb{N}} \tau_j B\right)
\le \sup_{j\in\mathbb{N}} \dim_{H}(\tau_j B)
= \dim_{H} (B)
&lt; \dim_{H}(A),$$
which is a contradiction.
The Hausdorff dimension has the following relevant properties:


Countable stability: Let $\{ E_j \}_{j\in \mathbb{N}}$ be a countable collection of sets.  Then
$$ \dim_H\left( \bigcup_{j\in\mathbb{N}} E_j \right) \le \sup_{j\in\mathbb{N}} \dim_{H}(E_j).$$
Monotonicity: If $E \subseteq F$, then $\dim_H(E) \subseteq \dim_H(F)$.


Suppose that $A$ and $B$ are arbitrary subsets of some Euclidean space $\mathbb{R}^n$ such that
$$ \dim_{H}(B) &lt; \dim_H(A). $$
Assume for contradiction that there exists a countable collection of translations $\{\tau_j\}_{j\in\mathbb{N}}$ such that
$$ A \subseteq \bigcup_{j\in\mathbb{N}} \tau_j B.$$
That is, assume that $A$ may be covered by a countable number of copies of $B$.  Translation is an isometry, hence it preserves the Hausdorff dimension.  Thus $\dim_{H}( \tau_j B) = \dim_H(B)$ for all $j$.  The Haudsorff dimension is countably stable and monotonic, thus
$$ \dim_{H}(A)
\le \dim_{H}\left( \bigcup_{j\in\mathbb{N}} \tau_j B\right)
\le \sup_{j\in\mathbb{N}} \dim_{H}(\tau_j B)
= \dim_{H} (B)
&lt; \dim_{H}(A),$$
which is a contradiction.