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@@ -13,7 +13,7 @@ We first begin with a few basic identities on the size of sets. Then, we will sh
 
 ## Proof that no. of available functions is greater than number of functions required to define the periodic function
 Consider any arbitrary periodic function in the interval $[-\pi,\pi]$. This can be represented as a series of values at various points in the interval. For example,  
-$ f(0) = ... , f(0.1) = ..., f(0.2) = ... $ and so on. At each point, we can assign any real number (i.e. $\in \mathbb R$). So, the number of possible periodic functions in an interval is of the order of $\mathbb R^{\mathbb R}$.
+$\qquad f(0) = ... , f(0.1) = ..., f(0.2) = ... $ and so on. At each point, we can assign any real number (i.e. $\in \mathbb R$). So, the number of possible periodic functions in an interval is of the order of $\mathbb R^{\mathbb R}$.
 --> don't quite remember how this goes.
 
 #<b>Why Fourier decomposition is plausible?</b>