# Variable derives itself

In Sipser's Introduction to the theory of computation (3rd edition), I found the following claim.

Consider the grammar:

\begin{align*} &R \to XRX \mid S \\ &S \to aTb \mid bTa \\ &T \to XTX \mid X \mid \epsilon \\ &X \to a \mid b \end{align*}

In this grammar, it holds that $$T \Rightarrow^* T$$.

Can anyone explain how this is true?

We say that $$\alpha \Rightarrow^* \beta$$ if $$\beta$$ can be derived from $$\alpha$$ in zero or more steps. (More fancily, $$\Rightarrow^*$$ is the reflexive-transitive closure of $$\Rightarrow$$.) In particular, it is always that case that $$\alpha \Rightarrow^* \alpha$$, for every $$\alpha$$, due to a derivation of zero steps.
In contrast, in your case it doesn't hold that $$T \Rightarrow^+ T$$. (This is derivation in one or more steps, or the transitive closure of $$\Rightarrow$$.)