Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
7  
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. –  Logan Maingi Oct 7 '12 at 20:23
1  
@LoganMaingi yes, I know that, when I said multiplication I meant the additive operation. –  user42912 Oct 7 '12 at 20:32
2  
@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. –  kahen Oct 7 '12 at 20:32
add comment

4 Answers

Addition is continuous because of the familiar 'addition law for limits':

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

share|improve this answer
add comment

In order to show that addition is continuous you're going to have to use the definition of continuity. I'm not sure which definition of continuity your book/class/notes is using, but once you write down the details of the definition it should be relatively clear. If not, you can edit your question to explain where you got stuck.

share|improve this answer
add comment

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.

share|improve this answer
add comment
up vote 1 down vote accepted

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.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.