One sample - why can't $\sigma^2=0$? The assumption is that
$Y_1,...,Y_n$ are independent and $Y_i \sim N(\mu,\sigma^2)$. The unknown parameters are $(\mu,\sigma^2) \in \mathbb R \times(0,\infty) $.
Question: Why can't variance be 0? meaning: $\sigma^2 =0$.
Kind regards,
 A: In normal distribution 
$$E(Y_i -\mu)^2=\sigma^2$$ 
so if $\sigma^2\rightarrow 0^+$ so $$E(Y_i -\mu)^2\rightarrow 0^+$$ in hence
$$P(Y_i=\mu) \rightarrow 1$$ so $Y_i$ are degenerate random variable in $\mu$. When variance equal 0 so the distribution is a degenerate  one. 
 R code, simulation
 sigma<-c(.1,.01,.001,.0001,.00001)
 n<-10000
 mat<-matrix(0,ncol=n,nrow=length(sigma))
 for( i in 1:length(sigma)){
 set.seed(i)
 mat[i,]<-rnorm(n,0,sigma[i])
 }

 epsilon<-.01 
 > length(which(abs(mat[1,])>epsilon))/n
 [1] 0.9185
 > length(which(abs(mat[2,])>epsilon))/n
 [1] 0.3168
 > length(which(abs(mat[3,])>epsilon))/n
 [1] 0
 > length(which(abs(mat[4,])>epsilon))/n
 [1] 0
 > length(which(abs(mat[5,])>epsilon))/n
 [1] 0

if you look at density of normal, at point $x=\mu$
$$f(\mu)=\frac{1}{\sqrt{2\pi \sigma^2}}$$ by  $\sigma \rightarrow 0^+$ 
$$f(\mu) \rightarrow \infty$$. on the other hand density at point $x=\mu$ is infinity . Other point be zero
  dnorm(0,0,0)
  [1] Inf
  > dnorm(1,0,0)
  [1] 0

at last note $\sigma^2$ can not be zero!since $f(x)$ is not defined at $\sigma^2=0$. just we can $\sigma^2 \rightarrow 0^+$.   we can calculate limiting distribution when  $\sigma^2 \rightarrow 0^+$. if $\sigma \rightarrow 0^+$ , the limiting distribution is degenerate distribution , and so it is not normal!. normal is continues and degenerate is discrete. So you have not a normal distribution with zero variance. 
A: You say in your question that the distributional form is an assumption (and therefore so is the allowable set of parameter values), so it is really up to you what assumption you wish to make.  It is possible to extend the family of normal distributions to include the case where $\sigma=0$, and in this case the distribution degenerates to a point-mass distribution on $\mu$.$^\dagger$  In answer to your follow up question in the comments, yes, it is possible to simulate this in R, and the standard commands for the norm distribution accomodate this case.
#Example of simulations from normal distribution with zero variance
mu    <- 10;
sigma <- 0;
rnorm(20, mu, sigma);

[1] 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10

There is no particular reason that you have to exclude the possibility that $\sigma = 0$ from your analysis.  Obviously if you exclude this case by assumption, then by assumption this outcome cannot hold.  However, you can certainly include that case as an allowable distribution in the analysis, in which case you are looking at a broader version of the family of normal distributions that includes this point-mass distribution.

$^\dagger$ It is a matter of convention whether or not the point-mass distribution is considered part of the family of normal distributions.  Most texts exclude it so as to focus on the continuous distributions, and retain certain general properties that don't hold with the inclusion of this case.  Nevertheless, it certainly would not be unreasonable to include it.  Whether or not you consider the point-mass distribution to be part of the "family of normal distributions", it is useful to include it in analysis for practical purposes, to ensure that the parameter space is closed.  Having a closed parameter space is useful for various reasons, including the fact that it ensures that in the case where you have observed data $y_1=\cdots=y_n$, your estimator $\hat{\sigma} = 0$ is in the parameter space.
