I have come across this post here and this question come to my mind:
"If $X=\{0,1\}^\Bbb N$, then $X$ is the set of all infinite sequences of zeros and ones, and the topology on $X$ is generated by finite sequences of zeros and ones. But no finite sequence belongs to $X$."
How, why??
A finite sequence on a set $S$ is by definition of map $f:\{1,2,\ldots,n\}\to S$, this is just a finite enumeration of some elements of $S$.
The elements of $X=\{0,1\}^{\Bbb N}$ are functions $f:\Bbb N\to \{0,1\}$, these are infinite sequences of $0$s and $1$s.
The topology that we can give $\{0,1\}^{\Bbb N}$ is the product topology of the discrete topology in $\{0,1\}$. Here we have projection maps $\pi_i:\{0,1\}^{\Bbb N}\to \{0,1\}$ that send $f$ to $f(i)$.
Thus a typical preimage $\pi_i^{-1}(0)$ is just the set of functions with $f(i)=0$ and $\pi_i^{-1}(1)$ is the set of functions with $f(i)=1$. A basic open set in the product topology is finite intersection $$V=\pi_{i_1}^{-1}(x_1)\cap \cdots \cap \pi_{i_k}^{-1}(x_k)$$
where $x_k$ is either $0$ or one. This is the set of functions for which $f(i_j)=x_j$ for $j=1,\ldots,k$.
Thus, if $O$ is an open set (a union of $V$s as above) and $f\in O$, $f$ is constrained to satisfy some conditions on finitely many if its values, since it will be in one of the $V$.
We want to understand how convergence works here. We already saw that if $f$ is a function in $\{0,1\}^{\Bbb N}$; a neighbourhood of $V$ is easily specified by a finite number of coordinates of $f$.Let $J$ be a finite subset of $\Bbb N$. Consider the neighbourhood $V_J(f)=\{g:g(j)=f(j),\;\forall j\in J\}$. These are seen to be open sets, and in fact they are a neighbourhood basis for at $f$. Since there are countably many finite subsets of $\Bbb N$, this is actually a countable neighbourhood basis. This means this space is first countable, so we may characterize its topology using sequences.
Now by the above, another function $g$ is close to $f$ if it coincides with $f$ in many yet only finite values. This space is also metrizable. Let's get an idea of what the metric should be, by looking at convergence. Suppose that $f_n\to f$ in $X$. This means that for each neighbourhood $V$ of $f$, all but finitely many $f_k$ fail to be in $V$. But a neighbourhood of $f$ is just a collection of functions that coincide with $f$ in finitely many places. In particular, we can look at the neighbourhoods $V_{\{1,2,\ldots,N\}}(f)$. Since $f_n\to f$, this means that for each $N$, all but finitely many of the $f_n$ agree with $f$ up to the $N$-th coordinate.
Note, for example, that the sequence $(1,0,\ldots)$, $(1,1,0,\ldots)$, $(1,1,1,0,\ldots)$, converges to $(1,1,1,1,\ldots)$.
Note that, in fact, the sets $V_{\{1,\ldots,k\}}(f)$ also form a basis of neighbourhoods at $f$: if $F$ is finite and $k$ is it's maximum element $V_{\{1,\ldots,k\}}(f)\subseteq V_F(f)$. This gives an idea on what our metric should be.
For $f,g\in X$, define $\Delta(f,g)$ to be the least element in $\{k:f(i)=g(i)\text{ for }1\leqslant i\leqslant k \}$.
If no least element exists, define this to be $+\infty$. Now define $d(f,g)=2^{-\Delta(f,g)}$. We would like to show that
PROP $f_n\to f$ in $X$ if and only if for every $\epsilon>0$, there exists $N$ such that $n\geqslant N\implies d(f_n,f)<\varepsilon$.
P Suppose first that $f_n\to f$, and pick $\varepsilon >0$. Now pick $N''$ such that $2^{-N''}<\varepsilon$. By the work done above, we know that there are only finitely many $f_n$ that do note belong to $V_{\{1,2,\ldots,N\}}$. Let $N'$ be such that $n>N'\implies f_n\in V_{\{1,2,\ldots,N\}}$. Let $N=\max\{N',N''\}$. Then $n>N$ gives $\Delta(f,f_n)>N$, so $d(f,f_n)<2^{-N}<\varepsilon$.
Now suppose that for any $\varepsilon>0$ there exists $N$ such that $n>N\implies d(f,f_n)<\varepsilon$. Let $V_F(f)$ be a basic open neighbourhood of $f$. Let $k$ be the largest number in $F$. Put $\varepsilon_F=2^{-k}$. Then there exists $N$ such that $d(f,f_n)<2^{-k}$, that is $k>\Delta(f,f_n)$. This means $f_n\in V_{\{1,\ldots,k\}}\subseteq V_F(f)$ if $n>N$, so $f_n\to f$. $\blacktriangle$
For any finite sequence ${\bf x}=(x_0,\dots,x_k)$, let $$G_{\bf x}\ :=\ \{ {\bf y}\in\{0,1\}^{\Bbb N}\ \mid\ y_i=x_i\ \text{ for all }\ i\le k \}\,.$$ This is a subset of $\{0,1\}^{\Bbb N}$, and these will form the basis of the topology.
