Convergence in the trivial topology

Question

I am currently trying to understand the terms neighbourhood and convergence. The definitions seems clear to me. I am however stuck seeing why this true:

In the trivial topology any sequence converges to every point (1).

By definition $x_n$ converges to $a$, if all neighbourhoods of $a$ contain every, but finitely many elements of $x_n$.

For example: Given $X = \{x,y\}, P(X) = \{\emptyset, \{x\}, \{y\}, \{x, y\}\} $ then $(X, P(X))$ is the trivial topology.

I may then define the constant sequeunce $x_n = x \ \forall n \in \mathbb{N}$

Obviously $x_n$ converges to $x$, but apparently also to $y$ (given the topology above), but I don't see why. $y$ has the neighbourhoods $N_1 = \{y\}$ and $N_2 := \{x,y\}$, obviously $x_n$ lies $N_2$, however not in $N_1$. Hence there is an open neighbourhood of $y$, that doesn't contain all but finitely many elements of $x_n$ and thus $x_n$ doesn't look like it converges to $y$.

The concepts are probably all very easy and intuitive, however I don't see where I go wrong and why (1) is in fact true.

Answer 1

The problem is that you are misunderstanding the meaning of “trivial topology”. It is the topology for which the (only) open sets are the whole space and the empty set. Given a set $X$, $\mathcal{P}(X)$ is the discrete topology, not the trivial one.

Answer 2

You're confusing the trivial topology with the discrete topology, which are very different: in the trivial topology every sequence converges, because the only nonempty open set is the whole set; in the discrete topology only eventually constant sequences converge.

The trivial topology on $X$ is $\{\emptyset,X\}$.

Answer 3

Trivial topology is $\{\emptyset, X\}$. Then obviously $x=x_n\to y$. Indeed, the only neighbourhood of $y$ is $\{x,y\}$ and all terms of the sequence $(x_n)$ belong to it.

The same argument applies to any $X\ne\emptyset$ with a trivial topology $\{\emptyset,X\}$.