Covering Dimension of a Subset of the Plane

Question

I am proving the equivalence of a few statements, and I'm stuck on one of them. I want to show that: If $K \subset [0,1]\times[0,1]$ is a non-degenerate continuum with empty interior, then $K$ has covering dimension $1$. (By a continuum, I mean a compact, connected, metric space. In this context, it is equivalent to a closed connected subset of $[0,1]\times[0,1]$.)

Outline of attempted proof: Let $K \subset [0,1]\times[0,1]$ be a non-degenerate continuum. Suppose $K$ has empty interior. Show that dim $K$ ≠ $0$. Show that dim $K$ ≠ $2$. We already know dim $K \leq$ dim $[0,1]$ + dim $[0,1]$ = $2$. Hence this shows that dim $K$ = $1$.

We know $K$ has empty interior and is closed. This means $K$ = $\partial K$. As $K$ is non-degenerate, it contains at least two points $a, b$. Suppose (with a view to contradiction) that it has dimension $0$. Then every open cover has a refinement with order at most $1$ - in other words, each open cover has a refinement such that all sets in it have empty intersection. Consider an open cover $\{U_n\}$ so that $a \in U_i, b \in U_j, i≠j$ and $a,b$ are not in any of the other sets in the cover. Now consider a refinement with order $1$. The existence of this refinement demonstrates that $a$ and $b$ are not connected, so $K$ is not a continuum. Thus dim $K$ ≠ $0$.

However, I am stuck on the second step. I'm unable to prove that the dimension of $K$ is not equal to $2$. Any help will be appreciated!

(If someone knows a theorem of a more general case, or results about dimension theory in Euclidean space, that would be fantastic! What I'm really after is the result, so a reference to a theorem or corollary in a book that directly proves my result is just as useful as an answer that contains a proof.)

Answer 1

This paper by Mardesic shows:

Let $X \subseteq \mathbb{R}^n$ which is compact and $k$-dimensional. Then there are coordinates $1 \le i_1 < i_2 \ldots i_k \le n$ such that $\operatorname{int}(\pi_{i_1, \ldots, i_k}[X]) \neq \emptyset$.

Now, your $K$ does not obey $\dim(K)= 0$ because that implies that $K$ is totally disconnected while it is connected. $\dim(K) =2$ is ruled out by the above theorem (or even on its quoted inspiration: a subset of $\mathbb{R}^n$ has dimension $n$ iff it has non-empty interior) as $K$ has empty interior.

So $\dim(K) = 1$ is forced.