Prove Addition is Continuous (without epsilon-delta!)

Question

Let $+ : \mathbb{R} \times \mathbb{R} \rightarrow \mathbb{R}$ be defined by $(x, y) \mapsto x + y$, and let both $\mathbb{R}$ and $\mathbb{R} \times \mathbb{R}$ have the usual topologies. My question is, is there a way to prove that $+$ is continuous without resorting to a $\varepsilon - \delta$ argument? I know it can be done easily (but, perhaps, tediously) with one, but it feels like this is close enough to the heart of the structure of $\mathbb{R}$ that there should be a more elegant way of doing it.

Answer 1

How about using Sequential criteria?

Let $$(x_n,y_n) \to (x,y)$$ Then $$x_n \to x, y_n \to y$$ Thus, $$x_n+y_n \to x+y$$ Hence, addition is continuous.

Answer 2

If $I$ and $J$ are open in $\mathbb R$ then $I\times J$ is open in $\mathbb R^2.$ To prove continuity of $f(x,y)=x+y,$ it suffices to show that when $f(x,y)\in K$, where $K$ is open in $\mathbb R,$ there exist open real $I,J$ such that $(x,y)\in I\times J$ and $f(I\times J)\subset K.$

There exists $r>0$ such that $(-r+f(x,y),r+f(x,y))\subset K.$ So let $I=(-r/2+x,r/2+x)$ and $J=(-r/2+y,r/2+y).$

We cannot avoid mentioning $r$. With a different topology on $\mathbb R,$ addition may fail to be continuous.

In general, $f:X\to Y$ is continuous iff whenever $f(p)\in V$ with $V$ open in $Y,$ there exists $U,$ open in $X$, with $p\in U$ and $f(U)\subset V.$ This could be called the topological generalization of the classical "$\epsilon$-$\delta$" definition of continuity.

Answer 3

I think if you want to avoid the $\varepsilon - \delta$ discussion then you should use the "preimage of open is open" definition of discontinuity. I just saw Mathmoore's answer so I guess my answer is really a rephrasing of his.

The topology on $\mathbb{R}$ is generated by open intervals. Hence it suffices to show that the preimage under addition of an open interval $(a, b)$ is an open subset of $\mathbb{R}^2$. The preimage is of course the set $\{(x, y)\in\mathbb{R}^2: a < x + y < b\}$. The solutions to the inequality $x + y < b$ is an open halfspace with base line given by the equation $x + y = b$. Similarly, the solutions to the inequality $a < x + y$ is an open subspace where the base line is given by the equation $x + y = a$ (paralel to the previous line!). Using that "finite intersections of open is open", or more simply that the intersection of the two open halfspaces is the open strip between the lines $x + y = a$ and $x + y = b$, we get that "$+$" is continuous (with respect to the standard topologies on $\mathbb{R}$ and $\mathbb{R}^2$.

Answer 4

This is not a rigorous answer.

By topological definition of continuity, we want to show that for every open subset $U$ of $\Bbb R$, the subset $+^{-1} (U)$ of $\Bbb R \times \Bbb R$ is also open.

So let us begin with $U \subset \Bbb R$ which is open. Take $t \in U$ arbitrarily.

Then $+^{-1}(t)=\{(y,t-y) : y \in \Bbb R\}$. That is inverse image of a point in $\Bbb R$ is a straight line in $\Bbb R \times \Bbb R$ with slope $-1$ and passes through the points $(t,0)$ and $(0,t)$. (Basically these are lines of the form $x+y=t$ in $\Bbb R^2$ and we are varying $t$ here.)

As $U$ is open, there exists an $r_t \gt 0$ such that the open interval $(t-r_t,t+r_t)$ is a subset of $U$.

So $+^{-1} ((t-r_t,t+r_t))=\{(y,p-y) \in \Bbb R \times \Bbb R: y \in \Bbb R, t-r_t\lt p \lt t+r_t\} \subset +^{-1}(U)$. Visualize this as a union of set of straight lines of the form $x+y=p$ where $t-r_t \lt p \lt t+r_t$. Try to prove it is an open set in $\Bbb R \times \Bbb R$ (Hint : Construct an open ball around every point of the set $+^{-1}((t-r_t,t+r_t))$ which is inside the set.)

Note that $\bigcup_{t \in U} +^{-1} ((t-r_t,t+r_t))=+^{-1}(U)$.

Answer 5

How about forgetting the definition of continuity and simply stating an elementary truth?

Let $x$ and $y$ be two numbers satisfying

$\tag 1 p \lt x \lt p + c$

$\tag 2 q \lt y \lt q + c$

with $c \gt 0$.

Then

$\tag 3 p + q \lt x + y \lt p + q + 2c$