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Let $(X,T)$ be a topological space, and let $f,g\colon X\to \mathbb R$ be continuous, real-valued functions.

Define the functions $f+g,f-g\colon X\to\mathbb R$ as follows:

$(f\pm g)(x)= f(x) \pm g(x)$ for every $x \in X$.

Prove these functions are continuous."

I know you can use proofs from Calc 1 to help determining continuity, but I'm not exactly sure what to do to get through this after some time.

Any help is greatly appreciated. Thank you.

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Addition is continuous. The composition of continuous functions is a continuous function. – copper.hat Mar 1 '13 at 8:00

Define $h : X \to \mathbb{R}^2$ by $h(x) = (f(x),g(x))$ Since $f,g$ are continuous we know that $h$ must be continuous (since each of it's components are continuous). We also know that the addition function $+ : \mathbb{R}^2 \to \mathbb{R}$ is continuous. Thus $(+ \circ h) = (f + g)$ is a continuous function from $X \to \mathbb{R}$ since a composition of continous functions is again a continuous function. The proof that $(f-g)$ is a continuous function is similar.

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To see that $h$ is a continuous function you can read the proof of proposition 2.6.29 in these notes… – Oliver E. Anderson Mar 1 '13 at 10:06

You need to know that $p: \mathbb{R}^2 \rightarrow \mathbb{R}$ defined by $p(x,y) = x + y$ is continuous. Suppose for now that this is a given.

Then for $f,g : X \rightarrow \mathbb{R}$ we can define the (diagonal) product map $f\nabla g: X \rightarrow \mathbb{R}^2$ that sends $x$ to $(f(x), g(x))$. This function is continuous whenever $f,g$ are because of the universal property of the product topology: a map into a product (here $\mathbb{R}^2$) is continuous iff all the compositions of the function with all the projections are continuous.

In this case $\pi_1 \circ (f\nabla g) = f$ and $\pi_2 \circ (f \nabla g) = g$, so the condition is satisfied, and $f \nabla g$ is thus continuous.

Now $f + g$ as defined is just $p \circ (f \nabla g)$, and thus a composition of continuous maps. The same works for minus, of course, mutatis mutandis.

To see that $p$ is continuous, we can use that $$|p(x,y) - p(u,v)| = |(x + y) - (u + v)| = |(x-u) + (y-v)| \le |x-u| + |y-v| \le 2\sqrt{(x-u)^2 + (y-v)^2}$$

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A more straightforward approach:

You need to show that the inverse image of every small open set is a small open set. You already know that for $f$ and $g$ (e.g. they are continuous).

Take a small neighborhood of, say, zero: $$A_n=\{x\big| \frac{1}{n} >|x|\}$$

Then what is $(f+g)^{-1}(A_n)$? It is all $x$s for which $\left|f(x) + g(x)\right| < \frac{1}{n}$.

All we really need now is to show that that is an open set. So, suppose that $x_0$ satisfies $|f(x_0) + g(x_0)| < \frac{1}{n}$. What you need to show is that (from the continuity of $f$ and $g$) there is a small neighborhood of $x_0$ for which $\left|f(x) + g(x)\right| < \frac{1}{n}$ holds.

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