# Let $\mathbb{R}^\mathbb{Z}$ be topological product, and assume that on $\mathbb{R}$ we have topology $T_{K}$

Let $$\mathbb{R}^\mathbb{Z}$$ be topological product, and assume that on $$\mathbb{R}$$ we have topology $$T_{K}$$ where $$T_{K} = \{ (a,b) : a,b \in \mathbb{R} , a,

$$K=\{ \frac{1}{n} : n \in \mathbb{N} \}$$ . Let $$(a_{n} )$$ be sequence in $$\mathbb{R}^\mathbb{Z}$$ , $$a_{n} (k) = k +\frac{1}{n}$$ for every $$k \in \mathbb{Z}, n \in \mathbb{N}$$. Is sequence $$(a_{n} )$$ convergent?

Any hint helps.

• $T_K$, as written, is not a topology. But the RHS of your def'n of $T_K$ is a base (basis) for a topology... BTW most (if not all) "typographic functions" in MathJax (LaTex) ,like \mathbb , don't require brace brackets when applied to a single key-stroke. So you can type \mathbb R instead of \mathbb {R}. And I found that \Bbb is exactly the same thing as \mathbb so you can type \Bbb R to get $\Bbb R$ – DanielWainfleet Apr 23 at 7:43

Claim: $$(a_n)$$ doesn't converge to anything.

proof: If $$x\neq(0,1,2,...)$$, it is easy to construct a neighboorhood that doesn't contain any $$(a_n)$$. For example, if the first coordinate $$x_1$$ is not equal to $$0$$, choose $$(a,b)$$ such that $$x_1\in(a,b)$$ and $$(a,b)\cap \{\frac{1}{n}\}_{n\in \Bbb N}=\emptyset$$. Let $$U=(a,b)\times \mathbb R \times \Bbb R \times...$$. It's clear that $$U$$ contains no $$(a_n)$$. Thus, we focus our attention on the case that $$x=(0,1,2,...)$$. Let $$U=\left\{(0,1)-K\right\}$$ $$\times$$ $$\Bbb R$$ $$\times$$ $$\Bbb R$$ $$\times...$$. We are again done.

(I believe everything becomes trivial if we view $$(a_n)$$ in this way)

$$a_1= (1,\quad 2,\quad3,\quad...)$$

$$a_2= (\frac{1}{2},\quad 1+\frac{1}{2},\quad2+\frac{1}{2},\quad...)$$

$$a_3= (\frac{1}{3},\quad 1+\frac{1}{3},\quad2+\frac{1}{3},\quad...)$$

.

.

.

$$a_n= (\frac{1}{n},\quad 1+\frac{1}{n},\quad2+\frac{1}{n},\quad...)$$

.

.

.