Theorem contradicting the necessity of Slutsky's theorem

Question

1. If $X_n\overset{p}{\rightarrow}X,Y_n\overset{p}{\rightarrow}Y$ then, show that $X_n+Y_n\overset{p}{\rightarrow}X+Y$.

I am not looking for the proof. My issue is, this seems contradictory to Slutsky's theorem, which is as follows.

2. If $X_n\overset{p}{\rightarrow}X,Y_n\overset{p}{\rightarrow}c\in\mathbb{R}$ then, $X_n+Y_n\overset{p}{\rightarrow}X+c$. [Replacing all convergences in distribution by convergence in probability as mentioned on Wikipedia]. Also, for $Y_n\overset{p}{\rightarrow}c$, $c$ being a constant is a necessity otherwise, the result is not guaranteed.

However, if theorem 1 holds then this necessity is useless. Slutsky's theorem should be true regardless of $c$ being a constant or a random variable. Where am I wrong? Is it because the necessity is only valid if we only have $X_n\overset{d}{\rightarrow}X$?

Answer 1

Indeed a note on Wikipedia writes "The theorem remains valid if we replace all convergences in distribution with convergences in probability".

The same could be said if you replace all convergences with almost sure convergence, or $L^p$ convergence. As far as I am concerned I find this note completely useless, as it could be misunderstood for something deeper you think you would have missed.