In Cantor set article, I presented the Cantor set which is a null set having the cardinality of the continuum. I present here a modification of the Cantor set named the **Smith-Volterra-Cantor set**.

### Construction of the Smith-Volterra-Cantor set

The **Smith-Volterra-Cantor set** (also named **SVC set** below) \(S\) is a subset of the real segment \(I=[0,1]\). It is built by induction:

- Starting with \(S_0=I\)
- \(S_1=[0,\frac{3}{8}] \cup [\frac{5}{8},1]\)
- If \(S_n\) is a finite disjoint union of segments \(s_n=\cup_k \left[a_k,b_k\right]\), \[S_{n+1}=\bigcup_k \left(\left[a_k,\frac{a_k+b_k}{2}-\frac{1}{2^{2n+3}}\right] \cup \left[\frac{a_k+b_k}{2}+\frac{1}{2^{2n+3}},b_k\right]\right)\]

And finally \(S=\displaystyle \bigcap_{n \in \mathbb{N}} S_n\). *The Smith-Volterra-Cantor set is created by repeatedly removing subintervals of width \(\frac{1}{2^{2n}}\) of the remaining intervals.* In comparison, the Cantor set is created by repeatedly removing the open middle third of the remaining intervals.

One can prove by induction that each closed subinterval of \(S_n\) has length \(l_n=\frac{2^n+1}{2^{2n+1}}\).

The Smith-Volterra-Cantor set is **closed** as it is an intersection of closed sets.

### The SVC set has measure \(\frac{1}{2}\)

At step \(n\) of the construction of SVC set, we remove \(2^n\) open disjoint intervals of length \(\frac{1}{2^{2n+2}}\). Hence, \(S\) is the set complement of an open set of measure \[\sum_{k=0}^\infty \frac{2^n}{2^{2n+2}}=\frac{1}{2}\] with respect to \([0,1]\). Which proves that \(S\) has Lebesgue measure equal to \(\frac{1}{2}\).

### The Smith-Volterra-Cantor set is nowhere dense

Let \(x,y\) be distinct points in \(S\), and assume without loss of generality that \(x < y\). We then chose a positive integer \(n\) such that \(l_n=\frac{2^n+1}{2^{2n+1}} < y-x\). Since the distance between \(x\) and \(y\) is greater than \(l_n\), \(x\) and \(y\) cannot lie in the same subinterval of \(S_n\). Because the subintervals that comprise \(S_n\) are closed and disjoint, there must be some point \(z \notin S\) between \(x\) and \(y\). Since the choices of \(x\) and \(y\) were arbitrary, it follows that between any two points in the SVC set there exists a point not in the set. Thus, \(S\) cannot contain any interval. Also, as \(S\) is closed, its complement is open. Therefore, we can find an open interval \(I\) such that \(z \in I \subset (x,y) \cap ([0,1] \setminus S)\). Which proves that \(S\) is nowhere dense.