Sequences and Nets: Does (1/n) Converge to [0,1]?

[ForMyThunder]

Say the real numbers were given a topology $$\left\{R,\phi, [0,1]\right\}$$. Does the sequence (1/n) converge to every point of [0,1] since it is a neighborhood of every point?

 

[Fredrik]

What is $R$ and $\phi$? Does your topology satisfy the definition of a topology?

 

[micromass]

If $$\mathbb{R}$$ has the topology $$\{\emptyset,[0,1],\mathbb{R}\}$$, then the sequence (1/n) converges to every point of $$\mathbb{R}$$!

 

[ForMyThunder]

If $$\mathbb{R}$$ has the topology $$\{\emptyset,[0,1],\mathbb{R}\}$$, then the sequence (1/n) converges to every point of $$\mathbb{R}$$!

Yeah, I guess you're right. Thanks.

 

[blue_raver22]

I can provide a mathematical explanation to this question. First, let's define what it means for a sequence to converge to a point in a topological space. A sequence (x_n) in a topological space X is said to converge to a point x in X if for every open set U containing x, there exists a natural number N such that x_n is in U for all n greater than or equal to N.

Now, let's consider the sequence (1/n) in the given topological space \left\{R,\phi, [0,1]\right\}. We can see that for any point x in [0,1], there exists an open set U containing x that is also a neighborhood of every point in [0,1]. So, by definition, the sequence (1/n) does converge to every point in [0,1].

However, it is important to note that this does not mean that the sequence (1/n) converges to [0,1] as a whole. In fact, it does not converge to any point outside of [0,1] since there is no open set containing points both inside and outside of [0,1].

In summary, the sequence (1/n) does converge to every point in [0,1] as it is a neighborhood of every point, but it does not converge to [0,1] as a whole in this topological space.