Almost sure convergence characterisations

According to Wikipedia https://en.wikipedia.org/wiki/Convergence_of_random_variables

$$X_n \to X$$ almost surely if and only if $$\forall\epsilon>0$$ $$P(\liminf_{n\to\infty}\{{\lvert X_n-X\rvert\lt \epsilon\}})=1$$

While, according to Almost sure convergence and lim sup

$$X_n \to X$$ almost surely if and only if $$\forall\epsilon>0$$ $$P(\limsup\limits_{n\to \infty}\{|X_n-X|\geq \epsilon\})=0$$

This seems to make sense to me because the events in the brackets are complements. However I am not sure if we had to swap $$\forall\epsilon>0$$ with $$\exists\epsilon>0$$

However my lecturer gave us this characterisation $$P(\cup_{m\ge1}\cap_{n\ge1}\cup_{k\ge n}\{|X_k-X|\gt\frac{1}{m}\})=0$$

Is the characterisation given by my lecturer correct? I am of the opinion that my lecturer's characterisation is $$\exists\epsilon>0$$

$$P(\limsup\limits_{n\to \infty}\{|X_n-X|\gt \epsilon\})=0$$.

My lecturer's characterisation seems correct intuitively. There is an epsilon such that for all n, there is a k greater than n such that $$|X_k-X|>\epsilon$$. This seems like divergence.

you don't need to switch $$\forall$$ and $$\exists$$, because given $$\varepsilon >0$$, those statements are, as you said, equivalent by taking complements; for fixed $$\varepsilon >0$$:$$1-P\left(\liminf\limits_{n\to \infty}\{|X_n-X|< \epsilon\}\right)=P\left(\limsup\limits_{n\to \infty}\{|X_n-X|\geq \epsilon\}\right)=0$$because $$\left(\limsup_{n\to \infty} A_n \right)^c = \liminf_{n\to \infty}A_n^c$$ for some events $$A_n$$, where $$A^c$$ means taking complements. So whenever one of those is fulfilled for all $$\varepsilon >0$$, the other one is, too.
The characterisation given by your lecturer is just the same; note that$$\bigcap_{n\ge1}\bigcup_{k\ge n}\left\{|X_k-X|\gt\frac{1}{m}\right\}=\limsup_{n\to \infty} \left\{ \vert X_k - X \vert > \frac{1}{m} \right\}.$$Now assume $$P\left(\bigcup_{m\ge1}\bigcap_{n\ge1}\bigcup_{k\ge n}\left\{|X_k-X|\gt\frac{1}{m}\right\}\right)=0$$ and take $$\varepsilon>0$$. Let $$l \in \mathbb{N}$$ be big enough, s.t. $$\frac{1}{l}<\varepsilon$$. Then $$0=P\left(\bigcup_{m\ge1}\bigcap_{n\ge1}\bigcup_{k\ge n}\left\{|X_k-X|\gt\frac{1}{m}\right\}\right)\overset{m=l}{\geq} P\left(\bigcap_{n\ge1}\bigcup_{k\ge n}\left\{|X_k-X|\gt\frac{1}{l}\right\}\right)\ge P\left(\limsup_{n\to\infty}\left\{|X_k-X|\gt\varepsilon\right\}\right).$$The other way round, if it holds for every $$\varepsilon>0$$, then it holds for all $$\frac{1}{m}$$ with $$m \in \mathbb{N}$$ and therefore $$P\left(\bigcup_{m\ge1}\bigcap_{n\ge1}\bigcup_{k\ge n}\left\{|X_k-X|\gt\frac{1}{m}\right\}\right)\leq\sum_{m=1}^\infty \underbrace{P\left(\bigcap_{n\ge1}\bigcup_{k\ge n}\left\{|X_k-X|\gt\frac{1}{m}\right\}\right)}_{=0}=0.$$