Can the Hausdorff dimension of a subset of the 2D plane be greater than 2? Although I've gotten the gist, I'm relatively unfamiliar with the bounds of a Hausdorff measure. From my understanding, very loosely speaking, it's a way of taking a geometric object, usually a fractal, and evaluating how many "degrees of freedom" an observer embedded within such object would have when it comes to traversing it's coordinate system.
E.g; Although the Sierpinski Triangle can be embedded into the 2D plane, it's does not have any area, yet it is more connected than a simple 1D curve (being compact and having infinite perimeter). As such, intuitively, its Hausdorff dimension is between 1, and 2 (evaluated at $\log_2 3\approx1.585$).

Can an object $\subset\mathbb{R}^2$ have Hausdorff dimension $>2$?

Looking at a List of fractals by Hausdorff dimension, the Hausdorff dimension of such objects seems to always be within $[0,2]$
Note: I'm simply an undergrad in mathematics, so an explanation at a lower level would be greatly appreciated.
 A: $\DeclareMathOperator{\diam}{diam}\DeclareMathOperator{\dim}{dim}$
No, this cannot occur.
First, let's quickly recall the definition and set some notation. For $X$ a metric space we have:

*

*A $r$-cover of $X$ is a countable set of open balls in $X$ each of radius $<r$, whose union is all of $X$.


*If $\mathcal{C}$ is a countable cover of $X$ by open balls and $a>0$, we let $\diam^a(\mathcal{C})$ be the sum of the $a$th powers of the radii of the balls in $\mathcal{C}$.


*For $d>0$, we set $\mathcal{H}^d(X)$ to be the limit, as $r\rightarrow 0$, of the infimum of $\diam^d(\mathcal{C})$ for $\mathcal{C}$ an $r$-cover of $X$.


*Finally, $\dim_H(X)=\inf\{d: \mathcal{H}^d(X)=0\}$.
The key point is that if $X$ is a subspace of $Y$ then every cover of $Y$ restricts to a cover of $X$, which in turn gives us $$\mathcal{H}^d(X)\le\mathcal{H}^d(Y)$$ for each $d$. Consequently we have $$\{d: \mathcal{H}^d(X)=0\}\supseteq \{d: \mathcal{H}^d(Y)=0\},$$ and this gives us the crucial result that dimension only increases when we add points: $$\color{red}{X\subseteq Y\implies \dim_H(X)\le \dim_H(Y).}$$
This immediately tells us that since $\mathbb{R}^2$ has Hausdorff dimension $2$, any subset of $\mathbb{R}^2$ has Hausdorff dimension $\le 2$.
