We begin with
Lemma 1.1. For all $c\in A$, the map $\tau_c: A\to A$ defined by $\tau_c(a)=c+a$ is a homeomorphism.
Proof. One can see that the restriction of the topology on $A\times A$ to $\{c\}\times A$ makes $\{c\}\times A$ homeomorphic to $A$ via the map $pr(c,b) = b$. Also, the map $add|_{\{c\}\times A}$ is continuous and bijective. This yields $\tau_c:=add|_{\{c\}\times A}\circ pr^{-1}: A\to A$ is also continuous and bijective. Its inverse map $\tau_{-c}$, by similar arguments, is also continuous. Hence $\tau_c$ is a homeomorphism (Q.E.D)
For most of cases, we want to equip topology for a given abelian group $A$. Here is a way to do this.
Definition. Let $A$ be a topological group, $x\in A$ be a point, then a fundamental system of open neighborhoods around $x$ is a set $\mathscr{N}_x$ containing open neighborhoods of $x$, such that for all $x\in U$, where $U\subset A$ open, there exists $V\in \mathscr{N}_x$ such that $V\subset U$. We denote a fundamental system of open neighborhoods around 0 as $\mathscr{N}$, instead of $\mathscr{N}_0$.
Lemma 1.2. The following holds for $\mathscr{N}$:
(T.G.1) $\forall U\in \mathscr{N}, \forall c\in U$, there exists $V\in \mathscr{N}$, such that $c+V\subset U$.
(T.G.2) $\forall U\subset \mathscr{N}$, there exists $V\in \mathscr{N}$, such that $V+V\subset U$.
Let $\mathscr{B}:=\{a+U|a\in A, U\in \mathscr{N}\}$, then $\mathscr{B}$ forms a base for the topology on $A$.
Proof. For (T.G.1), by Lemma 1.1, we have $U-c$ is also an open neighborhood of $0$, and hence, due to the definition of $\mathscr{N}$, there exists $V\in \mathscr{N}$, such that $V\subset U-c$, i.e. $V+c\subset U$.
For (T.G.2), let us look at $add^{-1}(U)$, it is an open neighborhood of $(0,0)$, and hence, there exists $V_1,V_2$: open neighborhoods of $0$ such that $add(V_1\times V_2)\subset U$. Take $V\in \mathscr{N}$, and $V\subset V_1\cap V_2$, we get $add(V+V)\subset U$.
Now, it follows by Lemma 1.1. that if $\mathscr{N}$ is a fundamental system around $0$, then $\{x+U|U\in \mathscr{N}\}$ is a fundamental system around $x$. Let $x\in A$ be any element, and $U_x$ is an open neighborhood of $x$, then $U_x-x$ is an open neighborhood of $0$, and by (T.G.1), there exists $V\in \mathscr{N}$, such that $V\subset U_x-x$, i.e. $x+V\subset U_x$, and $x+V\in \mathscr{B}$, and $x+V$ is an open neighborhood of $x$. This yields, any open subset of $A$ is a union of elements in $\mathscr{B}$. Hence, $B$ is a base for the topology on $A$. (Q.E.D)
We are now ready to see how to define a suitable topology for an abelian group so that it becomes a topological group.
Proposition 1.3. Let $\mathscr{N}$ be a set containing subsets of $A$, which contain $0$, such that $\mathscr{N}$ satisfies (T.G.1), (T.G.2) and
(T.G.3) $\forall U\in \mathscr{N}$, $-U\in \mathscr{N}$.
(T.G.4) $\forall U,V\in \mathscr{N}$, there exists $W\in \mathscr{N}$, such that $W\subset U\cap V$
then there exists a unique topology on $A$ that makes $A$ become a topological group, and $\mathscr{N}$ a fundamental system around 0.
Proof. We first define a topology on $A$ as follows. $U\subset A$ is open if either $U$ is empty, or for all $x\in U$, there exists $V\in \mathscr{N}$, such that $x+V\subset U$. With this kind of open sets, we now prove $A$ is a topological space.
i. $\emptyset, A$ are open.
ii. Let $U_i(i\in I)$ be open subsets of $A$, then it is obvious to see that $\cup_{i\in I} U_i$ is also open.
iii. Let $U_1,...,U_n$ be open subsets of $A$, we will prove that $U_1\cap ...\cap U_n$ is also open. By induction, it is sufficient for us to prove the case $n=2$. If $U_1\cap U_2=\emptyset$, then there is nothing to prove. Otherwise, let $x\in U_1\cap U_2$, then there exists $V_i(i=1,2)$, such that $V_i\in \mathscr{N}$, and $x+V_i\subset U_i(i=1,2)$. By (T.G.4), we can choose $V\subset V_1\cap V_2$ and $V\in \mathscr{N}$, and this yields $x+V\subset U_1\cap U_2$. Hence, $U_1\cap U_2$ is also open.
We now prove that with this kind of topology, $A$ becomes a topological group, that means, we have to prove $add$ and $inv$ are continuous. First, let $U\subset A, U\ne \emptyset$ be an open subset, then $inv^{-1}(U)=-U$. Take any $x\in -U$, we have $-x\in U$, and there exists $V\in \mathscr{N}$, such that $-x+V\subset U$, i.e. $x-V\subset U$. By (T.G.3), $-V$ is also in $\mathscr{N}$. This yields $-U$ is open, and $inv$ is continuous.
Next, we will prove that if $U\subset A, U\ne \emptyset$ is an open subset, then for any $c\in A$, $c+U$ is also open. Taking any $b\in c+U$, i.e. $b-c\in U$, and there exists $V\in \mathscr{N}$, such that $b-c+V\subset U$, i.e. $b+V\subset c+U$. And this yields $c+U$ is open.
Now, let $U\ne \emptyset, U\subset A$ be an open subset, consider $(c,d)\in add^{-1}(U)$, i.e. $c+d\in U$, then there exists $V\in \mathscr{N}$, such that $c+d+V\subset U$. By (T.G.2), there exists $W\in \mathscr{N}$ such that $W+W\subset V$, and this yields $(c+W)+(d+W)\subset U$. And by our previous argument, both $c+W, d+W$ are open and $(c+W)\times (d+W)\subset add^{-1}(U)$. Hence, $add$ is also continuous. Hence, $A$ is a topological group with this topology.
By (T.G.1), all elements in $\mathscr{N}$ are open subsets of $A$, and they form a fundamental neighborhood of $0$, since for any $U\subset A$: open, containing 0; there exists $V\in \mathscr{N}$, such that $0+V=V\subset U$. By Lemma 1.2, $\mathscr{B}:=\{a+U|a\in A,U\in \mathscr{N}\}$ forms a base for the topology on $A$, and this yields the uniqueness part. (Q.E.D)
We now come to the definition of topological rings.
Definition. Let $(A,+,.)$ be a commutative ring, then $A$ is said to be a topological ring if $A$ is also a topological space, and
i. (A,+) is a topological group.
ii. The map $mult: A\times A\to A$ defined by $mult(a,b)=ab$ is continuous.
Lemma 1.4. Let $A$ be a topological ring, and $a\in A$, then the map $m_c:A\to A$ defined by $m_c(a)=ac$ is continuous.
Proof. We have $\{c\}\times A$ is homeomorphic to $A$ via the map $pr(c,b)=b$, and the restriction of $mult|_{\{c\}\times A}$ is continuous, and hence, the composition $m_c=mult|_{\{c\}\times A}\circ pr^{-1}$ is continuous. (Q.E.D)
Similarly to earlier argument, we obtain
Lemma 1.5. Let $\mathscr{N}$ be a fundamental system around 0 of a topological ring $A$, then
(T.R.1) $\forall U\in \mathscr{N}, \forall c\in A$, there exists $V\in \mathscr{N}$, such that $cV\subset U$.
(T.R.2) $\forall U\in \mathscr{N}$, there exists $V\in \mathscr{N}$, such that $V.V:={uv|u,v\in V}\subset U$.
Proof. For (T.R.1), using Lemma 1.4, the map $m_c:A\to A$ is continuous, and this yields $m_c^{-1}(U)\subset A$ is also open, and contains $0$. So, there exists $V\in \mathscr{N}$, such that $V\subset m_c^{-1}(U)$, i.e. $cV\subset U$.
For (T.R.2), the map $mult: A\times A\to A$ is continuous, and hence, $mult^{-1}(A)$ is open in $A\times A$. Let $V_1\times V_2\subset mult^{-1}(U)$, and $V_1, V_2$ are open neighborhoods of $0$, we can now choose $v\in \mathscr{N}$ and $V\subset V_1\cap V_2$, and $V.V=mult(V\times V)\subset U$. (Q.E.D)
And similar to commutative group, here is a way to put a topology on a commutative ring $A$ so that it becomes a topological ring.
Proposition 1.6. Let $\mathscr{N}$ be a set containing subsets of $A$, which contain $0$, such that $\mathscr{N}$ satisfies (T.G.1)-(T.G.4) and (T.R.1), (T.R.2), then there exists a unique way to equip $A$ a topology such that $\mathscr{N}$ becomes a fundamental system around $0$.
Proof. By Proposition 1.3, there is a unique way to equip $A$ a topology such that $A$ becomes a topological group. So, it is sufficient to prove that, with this kind of topology, $A$ now becomes a topological ring. That means we have to check that the map $mult$ is continuous.
First, let $m_c: A\to A$ be defined as above, then for any $U\subset A$: open, non-empty, and $x\in m_c^{-1}(U)$, i.e. $cx\in U$. By (T.G.1), there exists $V\in \mathscr{N}$, such that $cx+V\subset U$, and by (T.R.1), there exists $W\in \mathscr{N}$, such that $cW\subset V$, and hence, $cx+cW\subset U$, that means, $x+W\subset m_c^{-1}(U)$. And this yields $m_c$ is continuous.
We will now prove that $mult$ is continuous. Let $U$ be as above, take $(c,d)\in mult^{-1}(U)$, i.e. $cd\in U$. By (T.G.1), there exists $V\in \mathscr{N}$ such that $cd+V\subset U$. And by (T.R.2), there exists $W\in \mathscr{N}$ such that $W.W\subset V$. Now, by (T.R.1), we can choose $W',W''\in \mathscr{N}$ such that $dW'\subset V, cW''\subset V$. By (T.G.4), there exists $W_c, W_d\in \mathscr{N}$ such that $W_c\subset W'\cap W, W_d\subset W''\cap W$, then
$$(c+W_c)(d+W_d)=cd+dW_c+cW_d+W_cW_d\subset U$$
Hence, $mult$ is a continuous map. This yields $A$ is a topological ring. (Q.E.D)
We now discuss about Cauchy sequence on a topological ring $A$.
Definition. Let $A$ be a topological ring with $\mathscr{N}$ is a fundamental system of open neighborhoods around $0$, and $(x_n)_n$ a sequence of elements in $A$, then $(x_n)_n$ is said to be a Cauchy sequence with respect to $\mathscr{N}$ if for every $U\in \mathscr{N}$, there exists an integer $n_U$ such that for all $m,n\ge n_U$, we have $x_m-x_n\in U$. We say that $x$ is a convergent value of $(x_n)_n$ if for every $U\in \mathscr{N}$, there exists an integer $n_U$, such that for all $m\ge n_U$, we have $x-x_m\in U$. We say $(A, \mathscr{N})$ is sequential complete with respect to $\mathscr{N}$ if any Cauchy sequence with respect to $\mathscr{N}$ converges in $A$.
Proposition 1.7. Let $A$ be a topological ring, and $\mathscr{N}$ is a fundamental system of open neighborhoods around $0$. If $\mathscr{M}$ is another fundamental system of open neighborhoods around $0$, then $(x_n)_n$ is a Cauchy sequence with respect to $\mathscr{N}$ iff $(x_n)_n$ is a Cauchy sequence with respect to $\mathscr{M}$. Hence, the definition of Cauchy sequences, and sequential complete do not depend on the choice of fundamental systems.
Proof. Assume that $(x_n)_n$ is a Cauchy sequence with respect to $\mathscr{N}$, then by definition, for any $V\in \mathscr{M}$, there exists $U\in \mathscr{N}$, such that $U\subset V$, then if we choose $m,n\ge n_U$, we get $x_m-x_n\in U\subset V$. Hence, $(x_n)_n$ is also a Cauchy sequence with respect to $V$. The converse is the same.
Assume that a Cauchy sequence $(x_n)_n$ with respect to $\mathscr{N}$ converges to $x\in A$, then it is clear from our earlier argument that $(x_n)_n$ also converges to $x$ with respect to $\mathscr{M}$. Hence, the definition of completion also does not depend on the choice of fundamental systems.
(Q.E.D)
We now discuss about Cauchy sequence on a topological ring $A$.
Definition. Let $A$ be a topological ring with $\mathscr{N}$ is a fundamental system of open neighborhoods around $0$, and $(x_n)_n$ a sequence of elements in $A$, then $(x_n)_n$ is said to be a Cauchy sequence with respect to $\mathscr{N}$ if for every $U\in \mathscr{N}$, there exists an integer $n_U$ such that for all $m,n\ge n_U$, we have $x_m-x_n\in U$. We say that $x$ is a convergent value of $(x_n)_n$ if for every $U\in \mathscr{N}$, there exists an integer $n_U$, such that for all $m\ge n_U$, we have $x-x_m\in U$. We say $(A, \mathscr{N})$ is sequential complete with respect to $\mathscr{N}$ if any Cauchy sequence with respect to $\mathscr{N}$ converges in $A$.
Proposition 1.7. Let $A$ be a topological ring, and $\mathscr{N}$ is a fundamental system of open neighborhoods around $0$. If $\mathscr{M}$ is another fundamental system of open neighborhoods around $0$, then $(x_n)_n$ is a Cauchy sequence with respect to $\mathscr{N}$ iff $(x_n)_n$ is a Cauchy sequence with respect to $\mathscr{M}$. Hence, the definition of Cauchy sequences, and sequential complete do not depend on the choice of fundamental systems.
Proof. Assume that $(x_n)_n$ is a Cauchy sequence with respect to $\mathscr{N}$, then by definition, for any $V\in \mathscr{M}$, there exists $U\in \mathscr{N}$, such that $U\subset V$, then if we choose $m,n\ge n_U$, we get $x_m-x_n\in U\subset V$. Hence, $(x_n)_n$ is also a Cauchy sequence with respect to $V$. The converse is the same.
Assume that a Cauchy sequence $(x_n)_n$ with respect to $\mathscr{N}$ converges to $x\in A$, then it is clear from our earlier argument that $(x_n)_n$ also converges to $x$ with respect to $\mathscr{M}$. Hence, the definition of completion also does not depend on the choice of fundamental systems.
(Q.E.D)
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