A question about the delta method in asymptotic distributions I am reading up on the delta method from its Wikipedia page. Under the heading Univariate delta method the statement of the method is as follows:
If
$$\sqrt{n}[X_n - \theta]\xrightarrow{\text{D}} \mathcal{N}(0,\sigma^2)$$ where $X_n$ is a sequence of random variables where $\theta$ and $\sigma$ are finite valued constants and $\xrightarrow{D}$ denotes convergence in distribution, then: $$\sqrt{n}[g(X_n) - g(\theta)]\xrightarrow{\text{D}} \mathcal{N}(0,[g^{'}(\theta)^2]\sigma^2)$$ Later on to prove this they ask us to note that $X_n \xrightarrow{P}\theta$ where $\xrightarrow{P}$ denotes convergence in probability
What justifies this claim? This seems to be sort of like a reverse central limit theorem and I feel like something very basic is in my blindspot. I need your help in figuring it out.

This basically follows directly from Slutsky's theorem. Below are the details.
By Slutsky's theorem, since $n^{-1/2} \to 0$ and $\sqrt{n} (X_n - \theta) \stackrel{D}{\to} N(0, \sigma^2)$, we have
\begin{equation*}
n^{-1/2} \left\{ \sqrt{n} (X_n - \theta) \right\} \stackrel{D}{\to} 0 \times N(0, \sigma^2).
\end{equation*}
Since this congergence in distribution to a constant implies convergence in probability to the same constant and since $X_n = n^{-1/2} \left\{ \sqrt{n} (X_n - \theta) \right\} + \theta$,
\begin{equation*}
 X_n \stackrel{p}{\to} 0 + \theta = \theta.
\end{equation*}
