Set of $\delta$s in $\epsilon-\delta.$

I've started learning real analysis in February of this year. By now, I'm able to prove most things in Calculus AB and BC from first principles but not much else. Hence, I am not too knowledgeable about the subject. I've recently tried proving harder things with just this knowledge. While writing these proofs, I often end up using the following object and I wish to know if there is a name for and/or there is any literature on it.

The $$\epsilon-\delta$$ definition of a limit states that, if for every $$\epsilon>0,$$ there exists a $$\delta>0$$ such that $$0<|x-a|<\delta\implies |f(x)-L|<\epsilon$$ where $$L$$ is some real number then $$\lim_{x \to a} f(x)=L.$$ Is there any name for the set of all $$\delta$$'s that work for some partiular $$\epsilon$$? Let $$\epsilon_I>0$$ be a constant. Then, is there a name for the set $$\Delta_{\epsilon_I}=\{\delta\mid 0<|x-a|<\delta\implies |f(x)-L|<\epsilon_I\}.$$

Here's the proof in which I used this set. It is a long one in my opinion. The theorem is: If $$f$$ is a continuous function on $$[a,b],$$ then the Darboux integral $$\int_a^b f$$ is defined.

Lemma 1: If $$G(r)=\sup\{f(x)\mid x \in [a,r]\}$$ where $$f$$ is a continuous function on $$[a,b]$$ (with $$a) and $$\text{Dom}(G)=(a,b),$$ then $$\lim_{r \to a^+} G(r)=f(a),$$ Proof: Since $$f$$ is continuous on $$[a,b],$$ we know that $$\lim_{x \to a^+} f(x)=f(a).$$ That is, for every $$\epsilon>0,$$ there must exist a $$\delta_3$$ such that $$0\le x-a<\delta_3 \implies |f(x)-f(a)|<\epsilon.$$ Rewriting the above yields $$x \in [a,a+\delta_3) \implies |f(x)-f(a)|<\epsilon.$$ This will come in handy shortly. Let $$\epsilon>0,$$ then we need to show that there exists a $$\delta_4$$ such that $$r \in (a,a+\delta_4) \implies |G(r)-f(a)|<\epsilon.$$ The above is just the $$\epsilon-\delta$$ definition written differently. Hence, if we choose $$\delta_3=\delta_4,$$ then we'll have $$r\in (a,a+\delta_3)$$ and $$x \in [a, a+\delta_3).$$ By definition, we have $$G(r)=\sup\{f(x)\mid x \in [a,r]\}.$$ By the extreme value theorem, there must exist some $$e\in [a,r]$$ such that $$f(e)=G(r).$$ However, we know that $$G(r)=f(e)\in \{f(x)\mid x \in [a,r]\}.$$ Since $$e\in [a,r]\subset [a,a+\delta_3),$$ it must be the case that $$x \in [a,a+\delta_3) \implies |f(e)-f(a)|<\epsilon.$$ Consequently, $$r\in (a,a+\delta_3) \implies |G(r)-f(a)|=|f(e)-f(a)|<\epsilon.$$ Hence, $$\lim_{r \to a^+} G(r)=f(a).$$ $$\square$$

Corollary 1: If $$g(r)=\inf\{f(x)\mid x\in [a,r]\}$$ with $$a and $$\text{Dom}(g)=(a,b),$$ then $$\lim_{r\to a^+}g(r)=f(a).$$ Proof: Consider the function $$-f(x).$$ Define $$G_1(r)=\sup\{-f(x)\mid x\in [a,r]\}$$ where $$\text{Dom}(G_1)=(a,b).$$ We know from lemma 1 that $$\lim_{r\to a^+} G_1(r)=-f(a).$$ It can be easily seen that $$G_1(r)=-g(r)$$ for all $$r\in (a,b).$$ Consequently, $$\lim_{r\to a^+} G_1(r)=-f(a)=\lim_{r\to a^+} -g(r)=-\lim_{r\to a^+}g(r).$$ Rearranging yields $$\lim_{r\to a^+} g(r)=f(a).$$ $$\square$$

Lemma 2: If $$G_2(r)=\sup\{f(x)\mid x\in[r,b]\}$$ where $$a and $$\text{Dom}(G_2)=(a,b),$$ then $$\lim_{r\to b^-} G_2(r)=f(b).$$ Proof: Consider the function $$f(-x+a+b)=f_1(x).$$ It can be proven that $$f_1$$ is continuous. We know that $$f_1(a+b-x)=f(x).$$ Additionally, we also know that $$x\in [r,b] \implies a+b-x\in [a,a+b-r].$$ Hence, we deduce that $$G_2(r)=\sup\{f(x)\mid x\in[r,b]\}=\sup\{f_1(a+b-x)\mid a+b-x\in [a,a+b-r]\}.$$ Then, define $$G_3(a+b-r)=\sup\{f_1(a+b-x)\mid a+b-x\in [a,a+b-r]\}.$$ Letting $$u=a+b-r$$ and $$v=a+b-x$$ yields $$G_3(u)=\sup\{f_1(v)\mid v\in [a,u]\}.$$ Taking the limit as $$r\to b^-$$ yields $$\lim_{r\to b^-} G_3(a+b-r)=\lim_{a+b-r\to a^+} G_3(a+b-r)=\lim_{u\to a^+} G_3(u).$$ Finally, applying lemma 1 to the last expression and un-doing some substitutions gives us $$\lim_{u \to a^+} G_3(u)=f_1(a)=f(-a+a+b)=f(b).$$ However, remember that $$G_3(a+b-r)=G_2(r).$$ Consequently, $$\lim_{r\to b^-} G_3(a+b-r)=\lim_{r\to b^-} G_2(r)=f(b).$$ $$\square$$

Corollary 2: If $$g_1(r)=\inf\{f(x)\mid x\in [r,b]\}$$ where $$a and $$\text{Dom}(g_1)=(a,b),$$ then $$\lim_{r \to b^-} g_1(r)=f(b).$$ Proof: Consider the function $$-f(x).$$ Define $$G_4(r)=\sup\{-f(x)\mid x\in [r,b]\}$$ where $$\text{Dom}(G_4)=(a,b).$$ Applying the result from lemma 2, we get $$\lim_{r\to b^-} G_4(r)=-f(b).$$ As was done in corollary 1, it can be deduced that $$g_1(r)=-G_4(r)$$ for all $$r\in (a,b).$$ Consequently, $$\lim_{r\to b^-} g_1(r)=\lim_{r\to b^-} -G_4(r)=-(-f(b))=f(b).$$ $$\square$$

With these results in hand, we can now proceed. Let $$r_0=a$$ and define $$H_i(r)=\sup\{f(x)\mid x\in [r_i,r]\},$$ $$h_i(r)=\inf\{f(x)\mid x\in [r_i,r]\}$$ where $$i\in \mathbb{Z}_{\ge 0}$$ such that $$r_i and $$\text{Dom}(H_i),\text{Dom}(h_i)=(r_i,b).$$ Given this, we will now create a procedure to find $$r_{i+1}.$$ Any $$r_k$$s found in this manner will be called "optimal $$r_k$$s for $$\epsilon_I$$." We will define $$\epsilon_I$$ in a minute. We know from lemma 1 and corollary 1 that $$\lim_{r\to r_i^+}(H_i(r)-h_i(r))=0.$$ Consequently, for every $$\epsilon_0>0,$$ there must exist a $$\delta_i$$ such that $$0 Let $$\epsilon_I>0$$ and let it be a constant. Then, define $$_i\Delta_{\epsilon_I}=\{\delta_i \mid 0 Next, let $$0 be a constant. Then, define $$\delta_{oi}=c\cdot \sup\{_i\Delta_{\epsilon_I}\}.$$ Now, we know that $$0 Finally, choose $$r_{i+1}=r_i+c\cdot \delta_{oi}.$$ Note that the same $$c$$ must be used to find out every $$r_{k}.$$

Now, let's say that this procedure (that is, all $$r_i$$s are optimal for $$\epsilon_I$$) is performed and we find that $$r_{i+1}=r_{i}+c\cdot \delta_{oi}\ge b$$ for some positive integer $$i.$$ In that case, simply choose $$r_{i+1}=b$$ and terminate the process. Our goal is to prove that this process always terminates for some $$i+1.$$

Lemma 3: $$r_i+\sup\{_i\Delta_{\epsilon_I}\}\le r_{i+1}+\sup\{_{i+1}\Delta_{\epsilon_I}\}$$ for all $$i\in \mathbb{Z}_{\ge 0}.$$ Proof: For all $$\delta_1,\delta_2>0$$ such that $$\delta_1<\sup\{_i\Delta_{\epsilon_I}\}$$ and $$\delta_2<\sup\{_{i+1}\Delta_{\epsilon_I}\},$$ we know that $$0 $$0 The absolute value signs aren't needed on the right side of $$\implies$$ here for obvious reasons. Rewriting this yields $$r_1\in (r_i,r_i+\delta_1) \implies H_i(r_1)-h_i(r_1)<\epsilon_I,$$ $$r_2\in (r_{i+1},r_{i+1}+\delta_2) \implies H_{i+1}(r_2)-h_{i+1}(r_2)<\epsilon_I.$$ Let's focus on the first statement above. Note that $$r_1' \in (r_i+\delta_1,r_i+\sup\{_i\Delta_{\epsilon_I}\})\implies H_i(r_1')-h_i(r_1')<\epsilon_I$$ since there must exist a $$\delta_1'$$ such that $$r_1' Because of this, we can just rewrite the two statements as follows: $$r_1\in (r_i,r_i+\sup\{_i\Delta_{\epsilon_I}\}) \implies H_i(r_1)-h_i(r_1)<\epsilon_I,$$ $$r_2\in (r_{i+1},r_{i+1}+\sup\{_{i+1}\Delta_{\epsilon_I}\}) \implies H_{i+1}(r_2)-h_{i+1}(r_2)<\epsilon_I.$$ Remember that $$H_i(r_1)=\sup\{f(x)\mid x\in [r_i,r_1]\},$$ $$h_i(r_1)=\inf\{f(x)\mid x\in [r_i,r_1]\},$$ $$H_{i+1}(r_2)=\sup\{f(x)\mid x\in [r_{i+1},r_2]\},$$ $$h_{i+1}(r_2)=\inf\{f(x)\mid x\in [r_{i+1},r_2]\}.$$ Next, assume that $$r_i+\sup\{_i\Delta_{\epsilon_I}\}> r_{i+1}+\sup\{_{i+1}\Delta_{\epsilon_I}\}.$$ We know that $$r_{i+1}>r_i.$$ Consequently, $$(r_{i+1},r_{i+1}+\sup\{_{i+1}\Delta_{\epsilon_I}\})\subset (r_i,r_i+\sup\{_i\Delta_{\epsilon_I}\}).$$ We want $$H_i(r_1')-h_i(r_1')<\epsilon_I,$$ $$H_{i+1}(r_2)-h_{i+1}(r_2)<\epsilon_I.$$ Hence, we can let $$r_1'\in (r_{i+1}+\sup\{_{i+1}\Delta_{\epsilon_I}\},r_i+\sup\{_i\Delta_{\epsilon_I}\})$$ to satisfy the first statement. Consequently, $$[r_{i+1},r_2]\subset [r_i,r_1']$$ for any $$r_2\in (r_{i+1},r_{i+1}+\sup\{_{i+1}\Delta_{\epsilon_I}\}).$$ By definition (of $$\sup\{_{i+1}\Delta_{\epsilon_I}\}$$), no other value of $$r_2$$ will satisfy the second statement above. Hence, it follows from our assumption that $$H_{i+1}(r_2)-h_{i+1}(r_2)\le H_i(r_1')-h_i(r_1').$$ But now let's see what happens if we let $$r_2'=r_1'.$$ It is still the case that $$[r_{i+1},r_1']\subset [r_i,r_1']$$ Hence, $$H_{i+1}(r_2')-h_{i+1}(r_2')\le H_i(r_1')-h_i(r_1').$$ However, we know that $$H_{i+1}(r_2')-h_{i+1}(r_2')\le H_i(r_1')-h_i(r_1')<\epsilon_I.$$ This is a contradiction since $$r_2'\not \in (r_{i+1},r_{i+1}+\sup\{_{i+1}\Delta_{\epsilon_I}\}).$$ Herefore, we can conclude that $$r_i+\sup\{_i\Delta_{\epsilon_I}\}\not > r_{i+1}+\sup\{_{i+1}\Delta_{\epsilon_I}\}.$$ Consequently, $$r_i+\sup\{_i\Delta_{\epsilon_I}\}\le r_{i+1}+\sup\{_{i+1}\Delta_{\epsilon_I}\}$$ as desired. $$\square$$

Next, assume that there always exists an $$r_i$$ such that $$r_{i+1}-r_i<\epsilon$$ where $$\epsilon>0.$$ Corollary 3: It follows from the above assumption that there must always be more than $$n+1$$ intervals after the procedure for any positive integer $$n.$$ Proof: If, at the end of this process, we had $$n+1$$ intervals, then it follows that there must be an $$r_k$$ such that $$r_{k+1}-r_k\ge r_{i+1}-r_i$$ for any $$i\in \mathbb{Z}_{\ge 0}.$$ However, this contradicts our assumption that $$r_{i+1}-r_i<\epsilon$$ for some $$i.$$ Hence, it follows from our assumption that there must be more than $$n+1$$ (for any $$n\in \mathbb{Z}_{>0}$$) intervals after the entire process.

Next, define $$\mathbb{o}=\{r_i\mid i\in \mathbb{Z}_{\ge 0}\}.$$ By construction, we know that $$\sup\{\mathbb{o}\}\le b.$$ Additionally, since $$c,\delta_{oi}>0,$$ it must be the case that $$r_{i+1}=r_i+c\cdot \delta_{oi}>r_i.$$ Consequently, $$\sup\{\mathbb{o}\}\not\in \mathbb{o}.$$

Lemma 4: $$r_i+\sup\{_i\Delta_{\epsilon_I}\}\le \sup\{\mathbb{o}\}$$ for all $$i\in \mathbb{Z}_{\ge 0}.$$ Proof: Assume that existed an $$i$$ such that $$r_i+\sup\{_i\Delta_{\epsilon_I}\}> \sup\{\mathbb{o}\}.$$ It then follows from lemma 3 that for all $$k\ge i,$$ we must have $$r_k+\sup\{_k\Delta_{\epsilon_I}\}\ge r_i+\sup\{_i\Delta_{\epsilon_I}\}>\sup\{\mathbb{o}\}.$$ It follows that $$0 Now, let $$z=r_i+\sup\{_i\Delta_{\epsilon_I}\}-\sup\{\mathbb{o}\}.$$ Hence, $$0 Since $$r_k<\sup\{\mathbb{o}\},$$ we get $$0 Note that since $$z$$ is a positive constant, we have found a lower bound for $$\sup\{_k\Delta_{\epsilon_I}\}.$$ That is, we have established $$0 for any $$k\ge i.$$ This, in turn, means that $$0 By definition, we know that there must be an $$r_k$$ arbitrarily close to (but not equal to) $$\sup\{\mathbb{o}\}.$$ Pick an $$r_k$$ with the property that $$\sup\{\mathbb{o}\}-r_k

We know that $$r_{k+1}=r_k+c^2\cdot\sup\{_k\Delta_{\epsilon_I}\}.$$ Consequently, $$\sup\{\mathbb{o}\}-r_{k+1}=\sup\{\mathbb{o}\}-r_k-c^2\cdot \sup\{_k\Delta_{\epsilon_I}\} Therefore, we have $$\sup\{\mathbb{o}\}-r_{k+1}<0.$$ Consequently, $$\sup\{\mathbb{o}\} This is clearly a contradiction. This means that our assumption was incorrect. Hence, $$r_i+\sup\{_i\Delta_{\epsilon_I}\}\le \sup\{\mathbb{o}\}$$ for all $$i\in \mathbb{Z}_{\ge 0}.$$ $$\square$$

Using the thought process from lemma 3, we know that $$r_3\in (r_i,r_i+\sup\{_i\Delta_{\epsilon_I}\}) \implies H_i(r_3)-h_i(r_3)<\epsilon_I.$$ Additionally, from lemma 4, we know that whatever $$r_3$$ might be, it has to be the case that $$r_3<\sup\{\mathbb{o}\}.$$ Equivalently, we could say that $$H_i(\sup\{\mathbb{o}\})-h_i(\sup\{\mathbb{o}\})=\sup\{f(x)\mid x\in [r_i,\sup\{\mathbb{o}\}]\}-\inf\{f(x)\mid x\in [r_i,\sup\{\mathbb{o}\}]\}\not< \epsilon_I$$ for any $$i\in \mathbb{Z}_{\ge 0}.$$ However, we know that there must always be an $$r_i$$ arbitrarily close to $$\sup\{\mathbb{o}\}.$$ The fact that no $$r_i$$ satisfies this means that lemma 2 and corollary 2 are violated. To see how, consider the function $$j(x)$$ with the property that $$j(x)=f(x)$$ when $$x\in [a,\sup\{\mathbb{o}\}].$$ Also consider the function $$J(r_i)=\sup\{j(x)\mid x\in [r_i,\sup\{\mathbb{o}\}]\}-\inf\{j(x)\mid x\in [r_i,\sup\{\mathbb{o}\}]\}.$$ By lemma 2 and corollary 2, we must have $$\lim_{r_i\to \sup\{\mathbb{o}\}^-} J(r_i)=0.$$ However, we just showed that no matter how close $$r_i$$ gets to $$\sup\{\mathbb{o}\}$$, $$J(r_i)-0$$ doesn't go below $$\epsilon_I.$$ This is a contradiction. Consequently, our process always guarentees $$n+1$$ intervals for some $$n \in \mathbb{Z}_{\ge 0}.$$ This means that our process always gives us a partition $$\mathcal{R}$$ such that $$a=r_0 Additionally, this partition also has the property that $$H_i(r_{i+1})-h_i(r_{i+1})<\epsilon_I$$ for any $$0\le i At last, we define $$u(f,\mathcal{R})=\sum_{i=0}^n H_i(r_{i+1})(r_{i+1}-r_i),$$ $$l(f,\mathcal{R})=\sum_{i=0}^n h_i(r_{i+1})(r_{i+1}-r_i).$$ Keep in mind that $$H_i,h_i$$ are functions and $$H_i(r_{i+1}),h_i(r_{i+1})$$ specific values that these functions take on. Subtracting the two equations yields $$u(f,\mathcal{R})-l(f,\mathcal{R})=\sum_{i=0}^n H_i(r_{i+1})(r_{i+1}-r_i)-\sum_{i=0}^n h_i(r_{i+1})(r_{i+1}-r_i)=\sum_{i=0}^n(H_i(r_{i+1})-h_i(r_{i+1}))(r_{i+1}-r_i).$$ Using the property that our partition has gives us $$u(f,\mathcal{R})-l(f,\mathcal{R})<\sum_{i=0}^n\epsilon_I(r_{i+1}-r_i)=\epsilon_I\sum_{i=0}^n(r_{i+1}-r_i)=\epsilon_I(b-a).$$ Hence, the difference between $$u(f,\mathcal{R})$$ and $$l(f,\mathcal{R})$$ can become arbitrarily small.

Theorem 1: If $$f$$ is a continuous function on $$[a,b],$$ then $$\int_a^b f$$ is defined. Proof: We already know that $$L_a^b(f) \le U_a^b(f)$$ since it is property 3 (and I have proven it earlier). To eliminate one of the remaining two cases, assume that $$L_a^b(f) By definition, $$u(f,\mathcal{R})\ge U_a^b(f),$$ $$l(f,\mathcal{R})\le L_a^b(f).$$ Consequently, $$l(f,\mathcal{R})\le L_a^b(f) Rearranging yields $$0\le L_a^b(f)-l(f,\mathcal{R}) Therefore, $$0 This implies that there exists a $$0 such that $$u(f,\mathcal{R})-l(f,\mathcal{R})>d.$$ This is a contradiction. Thus, for any continuous function $$f$$ defined on $$[a,b],$$ it must be the case that $$L_a^b(f)=U_a^f(f).$$ This means that $$\int_a^b f=L_a^b(f)=U_a^b(f)$$ is defined as desired. $$\square$$

• Well, as soon as one $\delta$ works, so do all smaller positive numbers. – Randall Nov 8 at 15:47
• Yeah, so it is essentially the set of $\delta$'s less than $\sup\{\Delta_{\epsilon_I}\}$ and greater than zero. Does it have a name though? Should I post the proof in which I used it here? – TheGeometer Nov 8 at 15:50
• I would be very interested in any proof which needs this specific set. Please edit your question to include this proof. – Carl Christian Nov 8 at 15:59
• I added the proof in which I used it. I didn't know uniform continuity at the time (and I still don't) but I wanted to figure it out on my own. Note that corollary 1/2 and lemma 1/2 are very similar statements. Either lemma 1 or 2 should give you the gist of the proof for the first two corollaries. – TheGeometer Nov 8 at 16:08
• In topology and real analysis, $U_{\epsilon}(x)$ is sometimes used to denote the set of elements $y$ satisfying $|y - x| < \epsilon$, and $f^{-1}(U_{\epsilon})$ would be used to denote the set of elements $z$ such that $f(z) \in U_{\epsilon}$. This is related to (but not of course not the same) as the set you describe. – Omnomnomnom Nov 8 at 16:17