Central limit theorem: Poisson equals Normal? Tell me where I'm wrong We just covered the Central Limit theorem in class, and I stumbled upon the following reasoning that makes me think I am missing some key part of... well, something. So, here goes:
Let $X_1$ and $X_2$ be independent Poisson random variables with parameters $\lambda _1$ and $\lambda _2$ respectively. Then, $X_1 + X_2 \sim P(\lambda_1 + \lambda_2)$. By induction, for any finite $n$, if $X_i \sim P(\frac1n)$ iid, $\sum ^n X_i= X \sim P(1)$, and this holds (I think) for $n \to \infty$.
But, by the CLT,
$$ \frac{\sum ^n X_i - n\cdot\frac1n}{\sqrt{n\cdot \frac1n}}\to \tilde Y \sim N(0, 1) $$
And rearranging I get $X \to Y \sim N(1, 1)$.
But $X$ was a Poisson random variable, therefore discrete, and thus its CDF is constant everywhere except over the integers; while $Y$ is a Normal, and therefore its CDF is never constant. How is one supposed to converge to the other?
This result is clearly wrong, but I can't understand why. Can anyone please explain where in my argument I made a mistake?

Suppose $X_1,X_2,X_3,\ldots\sim \operatorname{i.i.d.~Poisson(1)}.$ Then
$$
\frac{(X_1+\cdots+X_n)-n}{\sqrt n} \overset{\text{distribution}} \longrightarrow N(0,1) \text{ as } n\to\infty. \tag 1
$$
More generally, if $X\sim\operatorname{Poisson}(\lambda)$ then
$$
\frac{X-\lambda}{\sqrt\lambda} \overset{\text{distribution}}\longrightarrow N(0,1) \text{ as } \lambda\to\infty \\  \text{(where $\lambda$ is not constrained to take integer values).}
$$
But notice that in $(1),$ the distribution of each random variable $X_1,X_2,X_3,\ldots$ remains the same as $n\to\infty.$ You need that in order to apply the central limit theorem.
