Can a regular expression be infinite?

Problem Detail: 

I know that languages which can be defined using regular expressions and those recognisable by DFA/NFA ( finite automata ) are equivalent. Also no DFA exists for the language $\{0^n1^n|n \ge 0\}$. But still it can be written using regular expressions ( for that matter any non-regular language can be ) as $\{ \epsilon \} \cup \{01\} \cup \{0011\}......$ . But we know that every language that has a regular expression has a DFA that recognises it ( contradiction to my earlier statement ). I know this is a trivial thing, but does the definition of regular expression includes the condition that it should be finite ?

Asked By : sashas

Answered By : Ran G.

If regular expressions were allowed to be infinite, then any language would have been regular.

Given the language $L=\{w_1, w_2, \ldots\}$, we can always define the regular expression $R = w_1 + w_2 + \cdots$, which exactly defines $L$.
(Example: the regular expression $R_1 = \epsilon+0+1+00+01+10+11+\cdots$ defines $L_1=\{0,1\}^*$.)

We know that some languages are not regular, so this shows that infinite regular expressions describe a larger class of languages than finite regular expressions.

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Question Source : http://cs.stackexchange.com/questions/47835