A $\in$ $\tau$ iff for each $x$ $\in$ A $\exists$ $\epsilon \gt 0$ and a countable set $B$ such that (x- $\epsilon$, x+ $\epsilon$)\B $\subseteq$ A Let $X$ = $R$ and $\tau$ is a topology on $X$ where A $\in$ $\tau$ iff for each $x$ $\in$ A $\exists$ $\epsilon \gt 0$ and an at most countable set $B$ such that (x- $\epsilon$, x+ $\epsilon$)\B $\subseteq$ A. $\epsilon$ and $B$ are dependent on $x$.
a. Show $\tau$ is a topology:
Okay! So $R$ is in $\tau$ because, letting $\epsilon_x$ = any real number greater than 0, $B_x$ = $Q$, ($x-\epsilon_x,x+\epsilon_x$) \ $B_x$ $\subseteq$ $R$
$\emptyset$ $\in$ $\tau$ because the empty set is a subset of every set (Not sure about this explanation, but it's the only thing I can think of).
Now let U, V $\in \tau$. Then $\forall x \in U, \exists \epsilon_x$ and a countable set $B_x$ such that $(x- \epsilon_x,x+ \epsilon_x)\backslash B_x \subseteq U$.
Also, $\forall y \in V, \exists \epsilon_y$ and a countable set $B_y$ such that $(y- \epsilon_y,y+ \epsilon_y)\backslash B_y \subseteq V$.
Let $z \in U \bigcap V$. So $\exists \epsilon_z$ and a countable set $B_z$ such that $(z- \epsilon_z,z+ \epsilon_z)\backslash B_z \subseteq U$, and $\exists \epsilon_z$ and a countable set $B_z$ such that $(z- \epsilon_z,z+ \epsilon_z)\backslash B_z \subseteq V$. These follow from $z \in V$ and $z \in U$. So $\exists \epsilon_z$ and a countable set $B_z$ such that $(z- \epsilon_z,z+ \epsilon_z)\backslash B_z \subseteq U \bigcap V$.
So $U \bigcap V \in \tau$.
Let $\{V_i\}, i \in I$, be a family of members of $\tau$.
$\forall i \in I, \forall X_i \in V_i, \exists \epsilon_Xi$ and a countable set $B_Xi$ such that $(X_i- \epsilon_Xi,X_i+ \epsilon_Xi)\backslash B_xi \subseteq V_i$.
So the union of all $V_i$ will contain all the $\epsilon_Xi$ and $B_Xi$ from the original subsets, so the union of all $V_i \in \tau$. (Formally, I think I'd write: $\forall X_i \in \bigcup V_i, \exists \epsilon_Xi$ and a countable set $B_Xi$ such that $(X_i- \epsilon_Xi,X_i+ \epsilon_Xi) \subseteq \bigcup V_i$)
So $\tau$ is a topology (any comments on how this proof could be improved?)
b. Verify that 0 $\in$ closure((0,1)).
This one I wasn't sure, but I think I just say: Let $\epsilon \gt 0. (0- \epsilon, 0+ \epsilon)$ overlaps with (0,1) for every $\epsilon$. Therefore, 0 is in the closure. I'm pretty sure this is in good spirit but misses the mark.
c. Show that there is no sequence $\{X_n\}$ of (0,1) with $lim x_n$ = 0.
This is what I need help with proving at all. I supposed that an $x_n$ exists with limit 0. That means $\forall \epsilon \gt 0, \exists n_0 \in N$ such that $|x_n-0| \lt \epsilon \forall n \gt n_0$.
This means $x_n \lt \epsilon$ since $x_n$ is always positive. That's where I'm stuck.
Applying the topology, it's obvious that a dummy sequence such as 1/n wouldn't work because there is no countable subset that could be removed from tiny neighborhoods around each 1/n that would still be in R (or am I thinking wrong? The topology doesn't explicitly say the neighborhoods have to include the sequence points after the removal, so it could be Q.) I know I'm supposed to apply the topology somehow and arrive at a contradiction, but I'm not seeing how.
Thanks so much in advance!
 A: a. 


*

*$\mathbb{R} \in \tau$: correct, but it would be enough to take $B_x = \varnothing$.

*$\varnothing \in \tau$: wrong, you should use the definition: is it true that 
$$(\forall x \in \varnothing)(\exists \varepsilon > 0)(\exists B \subseteq \mathbb{R} \text{ countable}) \, (x-\varepsilon, x+\varepsilon) \setminus B \subseteq \varnothing \ ?$$
Yes, it is vacuously true, because in the beginning we have a for-all quantifier over the empty set.

*$U \cap V \in \tau$: wrong. You use the same symbols $\varepsilon_z, B_z$ twice for different things, which makes you implicitly (and incorrectly) assume they are equal. You should begin like this: let $z \in U \cap V$. 
Then there is $\varepsilon_1 > 0$ and countable $B_1 \subseteq \mathbb{R}$ such that $(z - \varepsilon_1, z + \varepsilon_1 ) \setminus B_1 \subseteq U$. 
There is also $\varepsilon_2 > 0$ and countable $B_2 \subseteq \mathbb{R}$ such that $(z - \varepsilon_2, z + \varepsilon_2) \setminus B_2 \subseteq V$.
Now do you see how to combine $\varepsilon_1, \varepsilon_2$ into some $\varepsilon > 0$ and $B_1, B_2$ into some countable $B \subseteq \mathbb{R}$ so that $(z - \varepsilon, z + \varepsilon) \setminus B \subseteq U \cap V$ ?

*$\displaystyle \bigcup_{i \in I} V_i \in \tau$ : your proof is chaotic and doesn't do what is required - you should again follow the definition. So we fix $\displaystyle x \in \bigcup_{i \in I} V_i$ and try to find $\varepsilon > 0$ and countable $B \subseteq \mathbb{R}$ such that $\displaystyle (x - \varepsilon, x + \varepsilon) \setminus B \subseteq \bigcup_{i \in I} V_i$. Since $\displaystyle x \in \bigcup_{i \in I} V_i$, there is some $j \in I$ such that $x \in V_j$. From the fact that $V_j$ is open and $x \in V_j$, we get an $\varepsilon > 0$ and countable $B \subseteq \mathbb{R}$ such that $(x-\varepsilon, x+\varepsilon) \setminus B \subseteq V_j$. But obviously $\displaystyle V_j \subseteq \bigcup_{i \in I} V_i$, so $\displaystyle (x-\varepsilon, x+\varepsilon) \setminus B \subseteq \bigcup_{i \in I} V_i$, which was our goal. 
b. In order to verify that $0 \in \operatorname{cl} (0, 1)$, you should fix an arbitrary neighborhood $U \in \tau$ of $0$ and show that $(0, 1) \cap U \neq \varnothing$. Considering the neighborhoods of form $(0 - \varepsilon, 0 + \varepsilon)$ isn't enough due to specific definition of the topology - there are more open sets than that.
Here is how you should begin: let $U \in \tau$ be a neighborhood of $0$. So there is $\varepsilon > 0$ and countable $B \subseteq \mathbb{R}$ such that $(-\varepsilon, \varepsilon) \setminus B \subseteq U$. Now if $U \cap (0, 1) = \varnothing$, then also $\big[ (-\varepsilon, \varepsilon) \setminus B \big] \cap (0, 1) = \varnothing$ so $(-\varepsilon, \varepsilon) \cap (0, 1) \subseteq B$. Is this possible?
c. What you are trying to do is very unclear to me. A sequence $x_n$ converges to $0$ is for each neighborhood $U \in \tau$ of $0$ there is $N \in \mathbb{N}$ such that $x_n \in U$ for all $n \geqslant N$. In other words, each neighborhood $U \in \tau$ of $0$ must contain all but finitely many terms of the sequence. I don't see why you are considering neighborhoods of $\frac{1}{n}$ and want them to be in $\mathbb{R}$ (?) after removing countably many points.
So here's how it should be done: fix any sequence $x_n$ of elements of $(0, 1)$. We want to show that $x_n$ does not converge to $0$, so we should point a neighborhood $U \in \tau$ of $0$ such that infinitely many terms of $x_n$ lie outside of $U$. Such a neighborhood can be of the form $(-\varepsilon, \varepsilon) \setminus B$ where $B \subseteq \mathbb{R}$ is countable. It's even possible to do it so that $(-\varepsilon, \varepsilon) \setminus B$ is disjoint from $\{ x_n : n \in \mathbb{N} \}$ - do you see how?
Response to comment: in topology open sets are understood to separate their elements from the outside. So if a point $p$ has an open neighborhood $U$ which separates a sequence $x_n$ from $p$, that is, infinitely many terms of $x_n$ lie outside $U$, then $x_n$ does not converge to $p$. That way the topology, by telling you which sets are open and which are not, establishes the notion of closeness.
The particular topology from the question admits open sets that separate their points from arbitrary countable sets, so it basically tells you that it takes more than countably many elements to converge to any point $p$. Since any sequence makes a countable set, it can not converge to $0$. 
More formally, let $x_n$ be an arbitrary sequence of elements of $(0, 1)$ and let $B = \{ x_n : n \in \mathbb{N} \}$, as you suggested. From the definition the set $U = (-1, 1) \setminus B$ is an open neighborhood of $0$ (check if you don't see it right away) and no term $x_n$ lies in $U$, thus $x_n$ does not converge to $0$. 
A: In this answer I give the standard axioms to define a local system of neighbourhoods $\mathcal{B}(x)$ for each $x$ that induces a topology in the way you are using.
Define, for $x \in \mathbb{R}$: $$\mathcal{B}(x) = \{B(x,\varepsilon, A): \varepsilon >0, A \subseteq \mathbb{R}, x \notin A, |A| \le \aleph_0 \}$$
where $$B(x,\varepsilon, A) = (x-\varepsilon, x+\varepsilon)\setminus A$$
We can check the axioms: 


*

*$\mathcal{B}(x) \neq \emptyset$, which is clear from the definition; e.g. $B(x,1,\emptyset) \in \mathcal{B}(x)$ for any $x$.

*$\forall B \in \mathcal{B}(x)$: $x \in B$, which is clear as for $B = B(x,\varepsilon, A)$ we have that $x \in (x-\varepsilon, x+\varepsilon)$ and $x \notin A$, so $x \in B(x,\varepsilon, A)$, as required.

*Suppose we have $B(x,\varepsilon_1, A_1)\in \mathcal{B}(x)$ and $B(x,\varepsilon_2, A_2)\in \mathcal{B}(x)$. Then define $\varepsilon = \min(\varepsilon_1, \varepsilon_2) > 0$ and $A  =A_1 \cup A_2$ (countable and misses $x$) and note that $B(x,\varepsilon, A) \subseteq B(x,\varepsilon_1, A_1) \cap B(x,\varepsilon_2, A_2)$ as required.

*Suppose that $U = B(x,\varepsilon, A) \in \mathcal{B}(x)$. Then define $V=U$ and suppose that $y \in V$. Then $y \notin A$ and $y \in (x-\varepsilon, x+\varepsilon)$. Find $\delta >0$ such that $(x-\delta, x+\delta) \subseteq (x-\varepsilon, x+\varepsilon)$ and note that $W = B(y,\delta, A)$ has the property that $W \subseteq U$, as required.
This neighbourhood system then gives a topology in the usual way: a set $O$ is open iff it is a neighbourhood of every one of its points iff it contains a member of $\mathcal{B}(x)$ for each $x \in O$, which is also how you defined it.
You always get a topology from a system obeying these axioms, in this way, and the $\mathcal{B}(x)$ are indeed local neighbourhood bases for $x$ in this topology.
It's clear that $0 \in \overline{(0,1)}$, for suppose $O$ is an open set that contains $0$, then $O$ contains some $B(0, r, A)$, and as $$O \cap A \supseteq (-r,r) \setminus A \cap (0,1) = (0,r)\setminus A \neq \emptyset $$
as all non-empty open intervals are uncountable and $A$ is only countable.
So $O$ intersects $(0,1)$, and $0$ is in its closure.
Suppose for a contradiction that we have a sequence $x_n \to 0$ where all $x_n \in (0,1)$. 
Then define $A = \{x_n \mid n \in \mathbb{N}\}$, whihc is countable and $0 \notin A$ as all $x_n > 0$. But then $B(0,1,A)$ is an open neighbourhood of $0$ that is disjoint from the sequence , so we cannot have $N$ such that for all $ n > N$ $x_n \in O$. So in fact $O$ shows that $x_n \not\to 0$, contradiction.
