Why the VC dimension to this linear hypothesis equal to 3? I am trying hard to understand this. Here is the scenario:

X = R^2
H = { h(x) = x + 10 }


I need to calculate the VC dimension for the above linear separator. Somehow, the VC dimension for this linear separator is 3. I just cannot understand how.

According to what I understand, VC dimension is the size of the largest finite subset of X which can be shattered by h. So if there exists a subset of size n where all points are shattered by h and there exists a subset of size n+1 where at least one point is not shattered by h, then the VC dimension will be n.


Is the VC dimension >= 1?



  Yes. We just need to pretend there is a single point on the line and by keeping the line (X-axis) steady we can flip which side is positive/negative



Is the VC dimension >= 2?



  Yes, because we could separate all 4 combinations { ++, --, +-, -+ } using a single line



Is the VC dimension >= 3?



  This should be NO, according to what I understand. How could we separate this case + - +?


But I was going through a video, that explains VC dimensions which says it is three and I do not understand how.
The VC dimension for the classifier depends on the dimension of space that your data points belong to.

In your problem, if your mean $x\in \mathbb{R^2}$, then the VC dimension is 3. For $\mathbb{R^2}$, you can always shatter any three general position points ("general position" means they do not coincidentally lie on the same line). For instance, consider three points (0,0), (0,1), (1,0). No matter how you assign labels to them, you can always seperate them by a line. If you consider point points with label (0,0), (0,1), (1,0), (1,1) with label [+,-,-,+], this is the "xor" case which cannot be seperated by a line.

For $x\in \mathbb{R}$, then the VC dimension is 2 because you cannot separate +-+.

In general for $x\in \mathbb{R^d}$, the VC dimension for a linear classifier is $d+1$.