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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.

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    $\begingroup$ It's a polynomial in two variables. $\endgroup$
    – Error 404
    Jul 4, 2017 at 4:58
  • $\begingroup$ Can you show that polynomials are continuous without an $\varepsilon - \delta$ argument somewhere along the way? $\endgroup$ Jul 4, 2017 at 5:00
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    $\begingroup$ I mean, sooner or later there will be an $\epsilon - \delta$ down there. The question is how far down you are willing to go... $\endgroup$
    – QC_QAOA
    Jul 4, 2017 at 5:05
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    $\begingroup$ Take a open set in $\mathbb{R}^2$, say $A$, $S_A=\{x+y: (x,y)\in A \}$ is also open set. $\endgroup$
    – MAN-MADE
    Jul 4, 2017 at 5:11
  • $\begingroup$ Do you know that rays form a subbasis of the topology on R? It suffices to show that their preimages are open. $\endgroup$ Jul 4, 2017 at 5:44

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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.

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    $\begingroup$ How do you justify you last "thus"? If I would have to prove that, I would use continuity of addition... $\endgroup$ Aug 9, 2017 at 11:03
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    $\begingroup$ @JensRenders That just follows from triangle inequality. $\endgroup$ Aug 9, 2017 at 15:14
  • $\begingroup$ Yes, it is an epsilon delta proof using the triangle inequality, and what you prove is continuity. Or how else do you prove the convergence of that last sequence $\endgroup$ Aug 9, 2017 at 15:27
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    $\begingroup$ @JensRenders $|x_n+y_n-(x+y)|=|x_n-x+y_n-y|\leq |x_n-x|+|y_n-y|$ $\endgroup$ Aug 9, 2017 at 15:46
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    $\begingroup$ @JensRenders If my definition of continuity is using $\epsilon-\delta$ and I show that "pre-image of open is open" is an equivalent criteria and then using that show "addition is continuous" then I'm using $\epsilon-\delta$ to prove "pre-image of open is open" implies continuity. The question required me to show continuity of addition without using $\epsilon-\delta$. It didn't say that I can't use a result which was proved using $\epsilon.$ $\endgroup$ Aug 9, 2017 at 16:53
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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.

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    $\begingroup$ An example to show why we can't avoid mentioning $r$: Let $\tau$ be the usual topology on $\mathbb R.$ Let $DO(\tau)$ be the set of $\tau$-dense members of $\tau$. (The set of dense open sets.). Let $X$ be the set $\mathbb R$ with the topology $\{\phi\}\cup DO(\tau)$. Then $f(x,y)=x+y$ is continuous in each variable separately, but $f:X^2\to X$ is not continuous. $\endgroup$ Aug 9, 2017 at 16:18
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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 $\mathbb{R}^2$.

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

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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

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

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