# Using the open set definition of continuity to directly prove a function is continuous

I'm looking to show that the function $s: \mathbb{R } \times \mathbb{R} \to \mathbb{R}$ where $s(x\times y) = x + y$ is continuous. In this case, the definition states if every open set $U$ of $\mathbb{R}$, the pre-image $s^{-1}(U)$ is an open set of $\mathbb{R}\times \mathbb{R}$.

The proof is very simple using projections $\pi_1 :X \times Y \to X$ and $\pi_2: X \times Y \to Y$, but that relies on the fact that the addition of two continuous functions is continuous. This is something that I've proven later in my set of questions.

So, how can I start with an open subset $U$ of $\mathbb{R}$ and somehow start to show that $s^{-1}(U)$ is open?

-
Hint: what is the preimage of an open interval? – Grumpy Parsnip Sep 20 '13 at 1:30
Just reword the $\epsilon-\delta$-proof of the continuity of $s$. – azarel Sep 20 '13 at 1:38
@GrumpyParsnip It's the set of all values that map to the open interval. So in this case, all points $(x,y)$ where $x+y$ is in the open interval. – MangoPirate Sep 20 '13 at 1:45
Can you show that that set is open? – Grumpy Parsnip Sep 20 '13 at 2:14
@GrumpyParsnip I've been working on some other stuff. I think we can do it by contradiction. If we assume the preimage is closed, then it has a maximum, $b$. Then $b+b \in U$. But then $b+b$ is the maximum of $U$ and since $U$ is open in $\mathbb{R}$, this is a contradiction. So the preimage is open. – MangoPirate Sep 20 '13 at 2:44

It is sufficient to show that $s^{-1}(a,b)$ is open for an open interval $(a,b)\subset\mathbb R$. Now $s^{-1}(a,b)=\{x\times y\, |\, a<x+y<b \}.$ This can be visualized as a diagonal strip in $\mathbb R^2$ of slope $-1$ with open boundaries hitting the $y$ axis at $a$ and $b$. Can you show this set is open?

-

Hint: Draw a picture of a basic open set in $\Bbb R^2$. Calculate its image under $+$.

-
I don't understand this hint. Isn't this a hint for how to show $s$ is an open map? – Grumpy Parsnip Sep 20 '13 at 3:48
@GrumpyParsnip: No. It's just that the situation for addition is rather simple. – dfeuer Sep 20 '13 at 11:48