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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<b\} \cup \{ (a,b) \backslash K : a,b \in \mathbb{R} , a<b \} $,

$K=\{ \frac{1}{n} : n \in \mathbb{N} \}$ . 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.

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  • $\begingroup$ $T_K$, 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 $\Bbb R$ $\endgroup$ – DanielWainfleet Apr 23 at 7:43
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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)

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

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

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

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

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