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Given a hypothesis class $H=\{h:X\to\{0,1\}\}$.

Let $c\in H$ be the correct predictor.

Denote $H^c = \{c\Delta h:h\in H\}$, where $c\Delta h=(h\backslash c)\cup (c\backslash h)$.

Please prove that VCdim($H^c$) = VCdim($H$)

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If some $A = (a_1,\dots,a_d\}\subset X$ is shattered by $H$, then, for any vector $b$ = $\{0,1\}^d$,$\exists$ $h \in H s.t. (h(a_1),\dots,h(a_d)) = b$.

It suffices to prove that for any vector $(h(a_1),\dots,h(a_d)) = b$, $\exists$ $h' \in H s.t. (c\Delta h'(a_1),\dots,c\Delta h'(a_d)) = (h(a_1),\dots,h(a_d))$.

Proof of my claim:

$\forall x_i \in A$, if $c(a_i) = 0$,let $h'(a_i) = h(a_i)$, then $c\Delta h' (a_i) = h(a_i)$.

If $c(a_i) = 1$, let $h'(a_i) = -h(a_i)$, then $c\Delta h'(a_i) = h(a_i)$.

Then $(c\Delta h'(a_1),\dots,c\Delta h'(a_d)) = (h(a_1),\dots,h(a_d))$.

For the same reason, the inverse is also true.

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