Are the order topology on the natural numbers and the discrete topology equivalent?

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I'm currently trying to get a better understanding of topology and I've read that the order topology on $\mathbb Z$ is equivalent to the discrete topology, since every subset of $\mathbb Z$ is open.

But isn't this the same for $\mathbb N$? We have the same cardinality and if we take the power set of $\mathbb N$ element subset is open as well. And if it isn't, why not?

Also I came across the following example:

$$\mathcal T_{\mathbb Z} := \{M\in \mathcal P(\mathbb Z): M = \emptyset \quad\text{or}\quad M = \mathbb Z\quad\text{or}\quad (-13 \in M \,\wedge 13 \notin M)\}$$

Doesn't this topology correspond to the discrete topology as well? I'm not really sure.