If $f$ and $g$ are uniformly continuous functions in $A$ show that $f+g$ is uniformly continuous in $A$.

Proof: because $f$ and $g$ are uniformly continuous on $A$ we can write

$$\forall\varepsilon>0,\exists\delta_1>0,\forall x,y\in A:|x-y|<\delta_1\implies|f(x)-f(y)|<\frac{\varepsilon}2$$ $$\forall\varepsilon>0,\exists\delta_2>0,\forall x,y\in A:|x-y|<\delta_2\implies|g(x)-g(y)|<\frac{\varepsilon}2$$

Then if I call $h(x)=f(x)+g(x)$ we can see that


and if I took $\delta_0=\min\{\delta_1,\delta_2\}$ then I can finally write

$$\forall\varepsilon>0,\exists\delta_0>0,\forall x,y\in A:|x-y|<\delta_0\implies|f(x)-f(y)|+|g(x)-g(y)|<\frac{\varepsilon}2+\frac{\varepsilon}2=\varepsilon$$

My question, it is the proof correctly written? Im unsure about the use of "$\frac{\varepsilon}2$" to hold the proof. Thank you in advance.


The idea is correct, but since you seem to learning to write out formal proofs I will add a few critiques:

  • You say that "we can see that $|h(x)-h(y)|$...", but this feels like it's missing justification. Just a brief nod towards the triangle inequality would go a long way towards distinguishing this writeup from that of someone who was just guessing, and in a truly formal setting you would make explicit use of the triangle inequality to justify this.

  • You are right to point out that introducing $\epsilon/2$ right from the beginning is a bit odd. It's very easy to deduce that from the formal definition, but in a fully-fleshed out proof this deduction should be explicit. Starting from $\epsilon > 0$ you know that $\epsilon/2 > 0$ and thus you can apply the uniform continuity definition to $\epsilon/2$. I believe this should be spelled out (of course, if this was a journal paper, I would apply a different standard of writing).

  • There is a small stylistic disconnect between the second-last line and the conclusion. In the second-last line you write "$\delta_0 = \min\{\delta_1, \delta_2\}$" which implicitly creates a context where you have already chosen a particular $\epsilon > 0$, since otherwise $\delta_1$ and $\delta_2$ don't exist. Then in the conclusion you use "$\forall \epsilon$", which feels like you've escaped that implicit context without actually saying so.

I believe ideally you should make this context very explicit: first explain that you are choosing an $\epsilon > 0$, then deduce the existence of $\delta_1, \delta_2$ corresponding to that choice of $\epsilon$ (this would be a more traditional place at which to introduce the $\epsilon/2$), then perform the calculation showing that $|h(x)-h(y)| < \epsilon$. Having closed that figurative bracket, you are then free to say $\forall \epsilon>0$ because the preceding argument works in generality.

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    $\begingroup$ This is the kind of answer I was waiting for. Thank you very much. $\endgroup$ – Masacroso Jun 1 '16 at 1:17

Yes, this is right! Scaling $\varepsilon$ in these arguments by a positive constant doesn't affect the conclusions. You really just have to think about this until you convince yourself. Instead of saying "for any $\varepsilon > 0$", maybe say "for any positive number". Then, given $\varepsilon > 0$, your "positive number" can be $\varepsilon$ itself, or $\varepsilon/2$, or $3\varepsilon$, ...

and a corresponding $\delta$ exists.

  • $\begingroup$ Yes, I understand. I can write the proof with words but my question is if the formal expression in symbols is correct, this is what I was afraid of. By your answer I understand that is right in the formal expression. $\endgroup$ – Masacroso Jun 1 '16 at 0:59

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