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Just wanted to verify if the following proof works: Suppose $f:X\rightarrow \mathbb{R}$ $g:X\rightarrow \mathbb{R}$ are continuous.

Want to show that: $h=f+g$ is continuous.

We are going to prove the statement using the localised definition of continuity: https://www.emathzone.com/tutorials/general-topology/continuity-in-topological-spaces.html

Let $x\in X$ and $U$ is open in $\mathbb{R}$ s.t $h(x)\in U$. Hence, $\exists \epsilon>0$ s.t $(h(x)-\epsilon,h(x)+\epsilon)\subset U$.

Since $f$ and $g$ are continuous at $x$ and $$f(x)\in B(f(x),\epsilon/2)$$ and $$g(x)\in B(g(x),\epsilon/2)$$ which are open sets in $\mathbb{R}$, this implies the $\exists V_1, V_2$ open in $X$ s.t $x\in V_1\cap V_2$ and $$f(V_1)\subset B(f(x),\epsilon/2) $$ and $$g(V_2)\subset B(g(x),\epsilon/2)$$

Hence, $\forall z\in V_1\cap V_2$ $$|h(z)-h(x)|\leq|f(z)-f(x)|+|g(x)-g(z)|< \epsilon$$ i.e $h(V_1\cap V_2)\subset(h(x)-\epsilon,h(x)+\epsilon)\subset U$.

Since $V_1\cap V_2$ is open in $X$ and $x\in V_1\cap V_2$, this implies $h$ is continuous at $x$.

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  • $\begingroup$ What does $+$ even mean? $\endgroup$
    – Randall
    Apr 16, 2019 at 14:23
  • $\begingroup$ I meant addition over $\mathbb{R}$ $\endgroup$
    – Jhon Doe
    Apr 16, 2019 at 14:24
  • $\begingroup$ The proof looks fine, except that $z$ should be $y$. $\endgroup$
    – Lee Mosher
    Apr 16, 2019 at 14:33
  • $\begingroup$ Thank you. Edited $\endgroup$
    – Jhon Doe
    Apr 16, 2019 at 14:34

1 Answer 1

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An alternative proof, if you know that $p: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ defined by $p(x,y)=x+y$ is continuous already:

Define $f \nabla g: X \to \mathbb R \times \mathbb R$ by $(f\nabla g)(x)=(f(x), g(x))$ and note that $f \nabla g$ is continuous as $\pi_1 \circ (f \nabla g) = f$ and $\pi_2 \circ (f \nabla g) = g$ are both continuous and $\mathbb{R}\times \mathbb{R}$ carries the product topology.

Then $f+g=h=p\circ (f \nabla g)$ is continuous as the continuous composition of continuous maps.

But your pointwise proof is fine too.

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  • $\begingroup$ How do we prove that the map $p$ is continuous? $\endgroup$ Apr 25, 2021 at 23:14
  • $\begingroup$ @GeorgeRevingston that’s an easy direct proof using the metric. And a standard fact about $\Bbb R$. $\endgroup$ Apr 26, 2021 at 6:10
  • $\begingroup$ Could you elaborate a bit please? Maybe give me a sketch of the proof? $\endgroup$ Apr 26, 2021 at 13:06
  • $\begingroup$ @GeorgeRevingston Given $(x,y) \in \Bbb R^2$ and $\varepsilon>0$, $(x-\frac{\varepsilon}{2}, x+\frac{\varepsilon}{2}) \times (y-\frac{\varepsilon}{2},y+\varepsilon}{2})$ is an open neighbourhood of $(x,y)$ that $p$ maps into $(p(x,y)-\varepsilon, p(x,y)+\varepsilon)$. This shows continuity st $(x,y)$. $\endgroup$ Apr 26, 2021 at 13:10

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