The real line with its additive group is a topological group?

Maybe it's a stupid question, I'm starting to study topological groups, I'm struggling to prove that the real line is a topological group with its additive group structure and Euclidean topology, precisely the part which we have to prove that if $g_1$ and $g_2$ $\in$ R, then the multiplication map $G\times G \to G$ defined by $m(g_1,g_2)=g_1 + g_2$ is continuous. Anyone can help me please.

• If you want to prove that the additive group of real numbers is a topological group, the relevant map is the addition map $(g_1,g_2) \mapsto g_1 + g_2$. The real numbers under multiplication are not even a group, because 0 is not invertible. Oct 7, 2012 at 20:23
• @LoganMaingi yes, I know that, when I said multiplication I meant the additive operation. Oct 7, 2012 at 20:32
• @LoganMaingi Actually you need to show that both $p(x,y)=x+y$ and $n(x)=-x$ are continuous. A clever little observation is that it's enough to show that $m(x,y) = x-y$ is continuous because $n(x) = m(0,x)$ and $p(x,y) = m(x,n(y))$ so if $m$ is continuous, then $n$ and $p$ are composites of continuous maps. Oct 7, 2012 at 20:32

Let $\mathbb R$ be the real line with its additive group structure and euclidean topology. We want to prove that the real line is indeed a topological group. Let $i:\mathbb R \to \mathbb R$, defined by $i(x)=-x$ and $m:\mathbb R \times \mathbb R \to \mathbb R$ defined by $m(x,y)= x+y$. we have to prove the following: (a) $i$ is continuous (b) $m$ is continuous.

(a) By a real analysis argument we know that $i$ is continuous, because $i$ is a polynomial function with real coefficients.

(b) We know that the projection $\Pi_1:X\times Y \to X$, $\Pi_1(x,y)=x$, where $X$ and $Y$ are topological spaces, is always continuous because for any open subset $U$ of X, we have $\Pi^{-1}(U)=U\times Y$ a open subset of $X \times Y$.

With the same argument the projection $\Pi_2:X\times Y \to Y$, $\Pi_2(x,y)=y$, where $X$ and $Y$ are topological spaces, is also continuous.

So by a real analysis argument which claims that the sum of continuous functions is continuous and as we know $m =\Pi_1 +\Pi_2$, then m is continuous.

Let $g_1,g_2 \in \mathbb{R}$ and $\epsilon>0$.

Then for each $(r_1,r_2)\in (g_1-\frac{\epsilon}{2},g_1+\frac{\epsilon}{2}) \times (g_2-\frac{\epsilon}{2},g_2+\frac{\epsilon}{2})$ we have that $$m(r_1,r_2)=r_1+r_2 \in (m(g_1,g_2)-\epsilon,m(g_1,g_2)+\epsilon) \ .$$

So if the definition for continuous functions you have is:

$f:A\longrightarrow B$ is continuous at $a \in A$ if for every open neighborhood $U$ of $f(a) \in B$ exist an open neighborhood $V$ of $a \in A$ such that $$\forall x \in V \; \Rightarrow\; f(x) \in U \ .$$

then $m$ is continuous.

$$\lim x_n=x \text{ and }\lim y_n=y\ \Rightarrow \lim (x_n+y_n)=x+y.$$