# Addition of continuous functions over topological spaces is continuous.

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 $$x$$.

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

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.
• How do we prove that the map $p$ is continuous? Apr 25, 2021 at 23:14
• @GeorgeRevingston that’s an easy direct proof using the metric. And a standard fact about $\Bbb R$. Apr 26, 2021 at 6:10
• @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 $(x,y)$. Apr 26, 2021 at 13:10