In binary classification, the VC-dimension is a measure of the capcity of a function class that can be used to derive error bounds for infinite function classes.

The VC-dimension of the set of classifiers that output the sign of a sine wave parametrized by a single parameter (the angular frequency of the sine wave) is infinite.

Another example of function class with infinite VC-dimension is the class of functions implemented by the nearest neighbor algorithm (the $K$-nearest neighbors algorithm using a single neighbor).

The plot shows the classifier computing the

possible labelings

with

This classifier has parameter

$\omega = $

We will show the following result on the VC-dimension of a set of classifier with a single parameter.

The set of classifiers $$ \F = \{f\ :\ f(x) = \sign\left(\sin(\omega x)\right),\ \omega \geq 0\}, $$ operating in one dimension with input space $\X = [0,2\pi]$, is infinite, i.e., $$ \VC(\F) = +\infty. $$

**Proof:**
Consider the labeled data set $\{(2\pi 10^{-i}, y_i)\}_{i=1}^n$ and choose, for any such data set, the parameter
$$
\omega = \frac{1}{2}\left( 1 + \sum_{i=1}^n \frac{1-y_i}{2} 10^i \right).
$$

**Step 1: correctly predicting negative labels**

The parameter can be rewritten as
$$
\omega = \frac{1}{2}\left(1 + \sum_{\{i\ :\ y_i=-1\}} 10^i \right).
$$
where we observe that, for any point $x_j =2\pi 10^{-j} $ in the considered data set such that $y_j=-1$, the term $10^{j}$ appears in the sum. This leads to
\begin{align*}
\omega x_j &= \pi 10^{-j}\left(1 + \sum_{\{i\ :\ y_i=-1\}} 10^i \right)\\
& = \pi 10^{-j}\left(1 + 10^j + \sum_{\{i\ :\ y_i=-1,\ i\neq j\}} 10^i \right)\\
&= \pi \left( 10^{-j} + 1 + \sum_{\{i\ :\ y_i=-1,\ i\neq j\}} 10^{i-j} \right)\\
&= \pi \left( 10^{-j} + 1 + \sum_{\{i \ :\ y_i=-1,\ i > j\}} 10^{i-j} + \sum_{\{i\ :\ y_i=-1,\ i < j\}} 10^{i-j} \right)
\end{align*}
For all $i>j$, the terms $10^{i-j}$ are positive powers of $10$ and thus are even numbers that can be written as $2k_i$ for some $k_i \in \N$. Therefore, we have
$$
\sum_{\{i \ :\ y_i=-1,\ i > j\}} 10^{i-j} = \sum_{\{i \ :\ y_i=-1,\ i > j\}} 2 k_i = 2 k
$$
for some $k\in \N$, which gives
\begin{align*}
\omega x_j &= \pi \left( 10^{-j} + 1 + \sum_{\{i \ :\ y_i=-1,\ i < j\}} 10^{i-j} \right) + 2k\pi .
\end{align*}
Regarding the remaining sum, we have
$$
\sum_{\{i\ :\ y_i=-1,\ i < j\}} 10^{i-j} < \sum_{i=1}^{+\infty} 10^{-i} = \sum_{i=0}^{+\infty} 10^{-i} - 1
$$
The geometric series $S_n = \sum_{i=0}^{n-1} 10^{-i}$, of first term $1$ and XXX raison XXX $0.1$, is known to converge towards
$$
\sum_{i=0}^{+\infty} 10^{-i} = \frac{1}{1-0.1},
$$
which gives
$$
\sum_{\{i\ :\ y_i=-1,\ i < j\}} 10^{i-j} < \frac{1}{1-0.1} -1 = \frac{1}{9}.
$$
Let define
$$
\epsilon = 10^{-j} + \sum_{\{i\ :\ y_i=-1,\ i < j\}} 10^{i-j}.
$$
and rewrite $\omega x_j$ as
$$
\omega x_j = \pi \left( 1 + \epsilon \right) + 2k\pi .
$$
Given that $10^{-j} \leq 0.1$ and the previous result, we have $0< \epsilon < 1/9 + 1/10 = 1$, and thus
$$
\pi < \pi(1+\epsilon) < 2\pi \quad \Rightarrow\quad \sin(\omega x_j) < 0.
$$
Hence, the classifier correctly predicts all negative labels $y_j = -1 = \sign(\sin(\omega x_j) ) = f(x_j)$.

**Step 2: correctly predicting positive labels**

The same steps can be reproduce with positive labels $y_j=+1$ with the difference that the term $10^j$ does not appear in the sum defning $\omega$. This leads to
\begin{align*}
\omega x_j &= \pi 10^{-j} \left( 1 + \sum_{\{i \ :\ y_i=-1,\ i \neq j\}} 10^{i} \right)\\
&= \pi \left( 10^{-j} + \sum_{\{i \ :\ y_i=-1,\ i > j\}} 10^{i-j} + \sum_{\{i \ :\ y_i=-1,\ i < j\}} 10^{i-j} \right)\\
& = \pi\epsilon + 2k\pi
\end{align*}
with
$$
0 < \pi \epsilon < \pi\quad \Rightarrow\quad \sin(\omega x_j) > 0.
$$
Thus, all positively labeled points are also correctly classified by $f$ using the particular choice of $\omega$. Since the steps above are valid for any labeling of the points, we proved that $\F$ shatters the set of points. In addition, the proof is valid for any number of points $n$, which shows that $\F$ can shatter sets of points of any size and thus has infinite VC-dimension.

We will show the following result on the VC-dimension for the $K$-nearest neighbors algorithm using $K=1$.

Let $\F$ denote the set of classifiers that are implemented by the $K$-nearest neighbors algorithm using $K=1$. Then, $$ \VC(\F) = +\infty. $$

To see this, simply observe that by definition the nearest neighbor algorithm assigns every point of the training set to its category: $f(x_i) = y_i$, whatever the particular value of $y_i$ is. Since this remains true for any training set size, we know that the algorithm can shatter a set of any size, simply by taking this set as the training set. Therefore, its VC-dimension cannot be bounded and is infinite.