Convergence and topology Please what is the classical method to answer this question, does the sequence converge in the given topology ? 
1) The sequence $\big(1+(-1)^n\big)_{n\in\mathbb N}$ in $(\mathbb{R},\tau)$ such that $\tau=\{U\subset \mathbb{R},\ 0\in U\}$.
2) The sequence $\big((\frac1n,1-\frac1n,\frac{n}{n+1})\big)_{n\in\mathbb{N}^*}$ in $X=[0,1]\times[0,1]\times[0,1]$ with the usual topology in $\mathbb{R}$.
I know that: $(u_n)$ converge to $\ell$ in a topological space $E$ if $$\forall V\in \mathcal{V}_\ell,\ \exists n_0\in\mathbb{N},\ \forall n,\ n\geq n_0\Rightarrow x_n\in V
$$
 but I don't know how to apply it !
Thank you
 A: Hint:
First you need to know what the open set looks like. 
For example, the open sets in Q1 is any subset of $\mathbb{R}$ containing $0$, for example $[0,2]$ is open. Then by definition, can you find any $x\in\mathbb{R}$ such that every open set containing $x$ containing tails of the sequence, hence your sequence is convergent? (ps, I think $\mathcal{T}$ you given is actually NOT a topology, you need to include $\emptyset$) 
Also note the topological space is not Hausdorff, so the limit may not be unique.

 for example $\forall x= 2$ is a limit.

For Q2, it's the product topology on $X=[0,1]^3$, you can use the fact that the sequence is convergent iff each component is convergent in $[0,1]$. OR, you can directly check by the definition.
A: The first sequence takes value either o 0r 2 depending on n is odd or even.clearly it can not converge other than o or 2,as you can always find an open set(take open set containing that point union 0) which will not contains points from the sequence.But this sequence can't converge to o as well as take open set[0,1)similarly sequence can not converge to 2 also.So first sequence do not converge in the given topology. 
A: (i) The sequence looks like: $2,0,2,0,2,0,2,0,\ldots$. A basic neighbourhood of a point $x$ is an open set (so any set that contains $0$, by definition) that also contains $x$. So clearly, every neighbourhood of $x$ contains $\{0,x\}$, and if it contains $\{0,x\}$, it's an open neighbourhood of $x$.
Now, the sequence converges to $x$ iff every neighbourhood of $x$ contains all but finitely $x_n$ from the sequence. So it cannot converge to any $x \neq 2$, as in that case the neighbourhood $\{0,x\}$ (which just equals $\{0\}$ for $x=0$) cannot contain any term $2$ from the sequence and this term occurs infinitely many times. 
But it does converge to $2$: $\{0,2\}$, the minimal neighbourhood contains all terms of the sequence, so this holds for all neighbourhoods of $2$ as well.
(ii) In the usual topology, $[0,1]^3$ has the product topology, so a sequence converges iff the different component sequences converge. This also gives you the limit.
