what does it mean to get the cdf of a constant variable I saw this theorem
If $X_n \ \xrightarrow{d}\ c$, where $c$ is a constant, then $X_n \ \xrightarrow{p}\ c$
In order to get $X_n \ \xrightarrow{d}\ c$, I need to prove $\begin{align}%\label{eq:union-bound}
   \lim_{n \rightarrow \infty} F_{X_n}(x)=F_X(x),
\end{align}$ which means I'm trying to get $F_c(x)$, my question is what does it mean to get the cdf of a constant variable and what would the density function for $c$ be? Is it just 0? If it is what is the intuition behind this?
 A: The cdf of a constant variable ($P(X=c)=1$) looks like this

Here 
$$F_c(x)=P(X\le x)=\begin{cases}1&\text{ if }& x\ge c\\ 0&\text{ otherwise}.\end{cases}$$
since even for $x=c$, $P(X\le c)=1$.
In this case there is no pdf. because there is no (normal) function 
whose integral from $-\infty$ to $x$ would look like $F_c(x).$
A: The cumulative distribution function for a constant random variable with $\mathbb{P}(X=c)=1$ is 
$$F(x) = \left\{\begin{matrix}0 &\text{ if }x < c \\ 1 &\text{ if }x \ge c    \end{matrix}     \right.$$
It does not have a meaningful density at $x=c$; if it did, then it would be something like a Dirac $\delta$-function 
A: There is nothing "special" finding the $\text{cdf}$ of a constant variable. Using the standard definition,
$$\text{cdf}_X(x)=\mathbb P(X\le x)=\mathbb P(c\le x)=\begin{cases}c> x\to 0,\\c\le x\to1\end{cases}$$
and the $\text{cdf}$ is the so-called Heaviside step function, with a shift, $H(x-c)$.

The density is a different matter. It is in principle the first derivative of the $\text{cdf}$, which is a discontinous function, hence it does not exist, strict sensu. In the sense of the theory of distributions (generalized functions, not probabilistic distributions !), it would be $\delta(x-c)$ (Dirac delta).

This is illustrated in the plot belwo, showing Gaussian distributions with decreasing standard deviations, where you can extrapolate convergence to Dirac delta and Heaviside step.

