Never a Best Response and Belief Consider a game in static, complete information environment, and the following definition of a strategy never being a best response to a player:


  A strategy $\sigma_i\in\Delta S_i$ is never a best response if there are no beliefs $\sigma_{-i}\in \Delta S_{-i}$ for player $i$ for which $\sigma_i\in BR_i(\sigma_{-i})$.


I want to understand the role of belief in this definition correctly. The definition of belief is given as:


  A belief of player $i$ is a possible profile of his opponent's strategies, $\sigma_{-i}\in \Delta S_{-i}$.


My understanding is:


  In situations where there is no strictly dominant strategy, a player asks herself "what do I think my opponents will do?". To resolve this uncertainty, the player assigns probability distribution over the set of pure strategies of opponent. One particular distribution produces a belief, one possible opponent's profile which is $\sigma_{-i}\in \Delta S_{-i}$.
  Now, I look at my simplex and say, is my particular strategy $\sigma_{i}\in \Delta S_{i}$ a BR to that of opponent? I strike out my $\sigma_i$ if I cannot find any probability distribution over the opponent's pure strategies such that $\sigma_i$ would be my BR.
  So, technically, I would have to compare my particular $\sigma_i$ against every possible probability distribution over the opponent's pure strategies to identify it as Not-BR, but in simple games, you often face with a situation where one of the pure strategies a player might have, say Column L of player 2, never gets underlined when you work out which ones for Player 2 is BR. This helps the process of eliminating all the strategies that are never a best response.


Is my understanding of belief and its role in finding the set of rationalizable strategies correct?
Your understanding is mostly correct, but the following sentence is not: 


  in simple games, you often face with a situation where one of the pure strategies a player might have, say Column L of player 2, never gets underlined when you work out which ones for Player 2 is BR.


Consider the following counterexample: 
\begin{array}{|c|c|c|c|}
\hline
 &amp;L&amp;C&amp;R\\\hline
U&amp;1,3&amp;3,4&amp;3,0\\\hline
M&amp;2,3&amp;0,1&amp;0,4\\\hline
D&amp;0,0&amp;1,2&amp;2,1\\\hline
\end{array}
In this game, $L$ is never a BR to any of player 1's pure strategies. However, if player 2 believes player 1 will play a mixed strategy that assigns $\frac12$ probability to $U$, $\frac12$ to $M$ and $0$ to $D$, then $L$ becomes a BR.