Making sense of this basis

Question

Could anyone help me make sense of how this basis works. I have $X = \bigcup_{n = 1}^\infty \mathbb{N}^n$. Now for $a \in \mathbb{N}^n$ and $A \subseteq \mathbb{N}^{n + 1}$, where $A$ is finite, we have $$U_a(A) = \{x \in X:\pi_{[1,n]}(x) = a,\pi_{[1,n+1]}(x) \notin A\}.$$ So, we have $$\{U_a(A):(\exists n \in \mathbb{N}) a \in \mathbb{N}^n, A \subseteq \mathbb{N}^{n + 1}\}$$ as a basis for a topology on $X$. I want to prove it is a basis on my own, but I still am having trouble getting an intuition of what these open sets exactly are.

To clarify, $\pi$ is a projection, i.e. $\pi_{[1,n]} (a_1,a_2,\cdots,a_{n},a_{n+1},\cdots) = (a_1,a_2,\cdots,a_n)$.

Answer 1

Regarding Tony's comment, I guess another hidden assumption in the definition of $U_a(A)$ is that $x$ must have length at least that of $a$, in order for $\pi_{[1,n]}(x)$ to be defined.

So for $a \in \mathbb{N}^n$ and $A \subseteq \mathbb{N}^{n+1}$, the set $U_a(A)$ contains all sequences $x$ of length at least $n$ such that the first $n$ components are the same as $a$, while the first $n+1$ components do not appear as an element of $A$.

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We need to check two properties of this collection in order to verify it is a basis. Do you see why the $U_a(A)$ cover $X$? For an arbitrary $x \in X$, you just need to find some $n \in \mathbb{N}$, some $a \in \mathbb{N}^n$ and some $A \subseteq \mathbb{N}^{n+1}$ such that $x \in U_a(A)$. For example, I believe $a=x_1$ and $A = \varnothing$ works.

For the other property, we need to show that if $U_a(A)$ and $U_{a'}(A')$ are sets in the collection, and $x$ lies in their intersection, then there exists $U_{a''}(A'')$ also in the collection such that $x \in U_{a''}(A'') \subset U_a(A) \cap U_{a'}(A')$.

There are some cases you need to handle.

• If $U_a(A) \cap U_{a'}(A') = \varnothing$, then there is nothing to prove.
• If $a=a'$, then consider $U_a(A \cup A')$.
• I think the only remaining case is $U_a(A) \subseteq U_{a'}(A')$ or $U_a(A) \supseteq U_{a'}(A')$, but I may be mistaken.

Answer 2

Maybe this can help you understand the sets $U_a(A)$.

For a given $A = \{\beta_1,\ldots,\beta_k\}$, $\beta_i = \begin{pmatrix}\beta_{i, 1}\\\vdots\\\beta_{i,n+1}\end{pmatrix}$ and $a=\begin{pmatrix}a1\\\vdots\\a_n\end{pmatrix}$ there are two cases. If $\exists i\in \{1,\ldots, k\}$ such that $\pi_{[1,n]}(\beta_i) = a$, then the set $U_a(A)$ has elements like $x=\begin{pmatrix}x_1\\\vdots\\x_m\end{pmatrix}$ ($m>n$) which satisfy $\begin{pmatrix}x_1\\\vdots\\x_n\end{pmatrix}=\begin{pmatrix}a1\\\vdots\\a_n\end{pmatrix}$ and at the same time, $x_{n+1}\not\in\{\beta_{i,n+1}:\pi_{[1,n]}(\beta_i)=a\}$. So to be $U_a(A)$ we require that the fist $n$ entries of $x$ should match $a$ and we also require that $x_{n+1}$ is not equal to $\beta_{i,n+1}$ for any of the betas which match with $a$, (i.e. $\pi_{[1,n]}(\beta_i)=a$).

If none of the $\beta_{i}$ match with $a$, i.e. $\forall k\in\{1,\ldots,k\}$ we have $\pi_{[1,n]}(\beta_i)\neq a$, then the set $U_a(A)$ is just all tuples whose first $n$ entries match $a$, $U_a(A)=\{x\in X:\pi_{[1,n]}(x)=a\}$. This automatically excludes $\pi_{[1,n+1]}(x)\in A$ since none of the betas match with $a$.

Answer 3

This is a topology for the string space S,
of all finite strings of positive integers N.

Let variables be finite strings and write ab for the
concantenation of a and b. If a = 34, b = 7, ab = 347.
Let z be the empty string. Let len a = unique n with a in N^n.
For a in S, A subset N, let B(a,A) = { axs : x in A, s in S }.

B = { B(a,A), {z} : a in S, A cofinite subset N } is a base for S.
This is the base you have using simple notation and with the
addition of {z} = $N^0$ to assure every string is in a base set.

Proof. Clearly the empty string is in a base set.
If a is not the empty string, then there is some b in S
and x in N with a = bx. Since x in cofinite A = N - {x+1},
a in B(b,A). Thus S is the union of the base sets.

Let B(a,A), B(b,D) be two base sets. In the case they
intersect, let s be a mutual point. Wlog len a <= len b.
If len a = len b, then a = b and s in B(a, A$\cap$ D)
which is a subset of both B(a,A) and B(b,D).
If len a < len b, then some x in A, u in S with axu = b.
Thus B(b,D) subset B(a,A).