What's the topological dimension of Sierpiński's Triangle?

Question

I am aware that Sierpiński's Triangle is a fractal, with Hausdorff dimension $1.5850$. Therefore my intuition leads me to believe it's topological dimension is 1 (as the topological dimension must be less than the Hausdorff dimension). However, since it's area is 0 this makes me believe it has topological dimension 0 since it's a collection of infinitely many points. Some clarification on this would be fantastic. Thank you.

Answer 1

The Sierpinski triangle contains a lot of line segments so I would think its dimension would be at least 1.

There's more than one definition of topological dimension but, generally, these are inductive definitions. Loosely, we might say that a set has dimension $n$, if there is a basis for its open sets whose boundaries have dimension $n-1$. The boundaries of the open balls in $\mathbb R^3$ are spheres and the surface of a shpere has dimension 2, for example.

Of course, we need a place to start so we might define $\dim(\varnothing)=-1$. This implies that the zero dimensional sets are exactly the totally disconnected sets. In particular, any finite set in Euclidean space has dimension zero.

Now, let $T$ denote the open triangle in the plane with vertices $$(0,0), \: (1,0), \text{ and } (1/2,\sqrt{3}/2).$$ We can obtain a basis for the open sets in the Sierpinski triangle by intersecting the Sierpinski triangle with images of $T$ under the application of functions generated by the IFS. Two such open triangles are shown in the figure below. Note that the boundary of these sets in the Sierpinski triangle are exactly the vertices of the smaller triangles. Thus, the basis sets have boundaries with dimension zero. So the Sierpinski triangle has dimension 1.