Consider the parameter space $\boldsymbol{\Theta}=\boldsymbol{\Theta_{1}}\times \boldsymbol{\Theta_{2}}$, which is a compact subset of $R^{n}$, and contains the true parameter vector $\boldsymbol{\theta^{*}}=(\boldsymbol{\theta_{1}'^{*}},\boldsymbol{\theta_{2}'^{*}})'$.

Consider also the parameter $\boldsymbol{\theta}=(\boldsymbol{\theta_{1}}',\boldsymbol{\theta_{2}')'}$, where $\boldsymbol{\theta_{1}}\in\boldsymbol{\Theta_{1}}\subset R^{n_{1}}$, $\boldsymbol{\theta_{2}}\in\boldsymbol{\Theta_{2}}\subset R^{n_{2}}$.

Define $\boldsymbol{\Theta_{r}}$ to be the restriction of the parameter space $\boldsymbol{\Theta}$ to $\boldsymbol{\Theta_{r}}=\boldsymbol{\Theta_{1}}\times N_{r}(\boldsymbol\theta_{2}^{*})$, where $N_{r}(\boldsymbol\theta_{2}^{*})=\{\boldsymbol\theta_{2}\in \boldsymbol\Theta_{2}:\lVert\boldsymbol\theta_{2}-\boldsymbol\theta_{2}^{*}\rVert\leq r\}$ for some small $r>0$.

Is it "topologically" correct to say that $\boldsymbol{\Theta_{r}}$ defines a local (nonzero measure) neighbourhood of the true parameter space?

I then want to consider a neighbourhood $B_{\epsilon}(\boldsymbol\theta^{*})\subset \boldsymbol{\Theta_{r}}$ with compact compliment $B_{\epsilon}^{c}(\boldsymbol\theta^{*})$ in $\boldsymbol{\Theta_{r}}$, and $\epsilon>0$, in order to prove $\limsup_{N\to\infty}max_{\boldsymbol{\theta}\in B_{\epsilon}^{c}(\boldsymbol\theta^{*}) }E[L_{N}(\boldsymbol\theta)-L_{N}(\boldsymbol\theta^{*})]<0$, where $L$ is my function.

For this to make sense as far as I can see I can't have $N_{r}(\boldsymbol\theta_{2}^{*})\subset B_{\epsilon}(\boldsymbol\theta^{*})$ because then $N_{r}(\boldsymbol\theta_{2}^{*})\cap B_{\epsilon}^{c}(\boldsymbol\theta^{*})=\varnothing$. So I guess what I really need is part of $N_{r}(\boldsymbol\theta_{2}^{*})$ to lie in $B_{\epsilon}(\boldsymbol\theta^{*})$ and part of it in $B_{\epsilon}^{c}(\boldsymbol\theta^{*})$. Is this "topologically" feasible?