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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 $a<b$) 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 $a<b$ 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 $a<b$ 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 $a<b$ 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 $r_i<b$ 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 $$0<r-r_i<\delta_i\implies |H_i(r)-h_i(r)|<\epsilon_0.$$ Let $\epsilon_I>0$ and let it be a constant. Then, define $$_i\Delta_{\epsilon_I}=\{\delta_i \mid 0<r-r_i<\delta_i\implies |H_i(r)-h_i(r)|<\epsilon_I\}.$$ Next, let $0<c<1$ be a constant. Then, define $$\delta_{oi}=c\cdot \sup\{_i\Delta_{\epsilon_I}\}.$$ Now, we know that $$0<r-r_i<\delta_{oi} \implies r\in (r_i,r_i+\delta_{oi})\implies |H_i(r)-h_i(r)|<\epsilon_I.$$ 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 $$0<r_1-r_i<\delta_1\implies H_i(r_1)-h_i(r_1)<\epsilon_I,$$ $$0<r_2-r_{i+1}<\delta_2\implies H_{i+1}(r_2)-h_{i+1}(r_2)<\epsilon_I.$$ 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 $r_1'<r_i+\delta_1'.$ 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 $$0<r_i+\sup\{_i\Delta_{\epsilon_I}\}-\sup\{\mathbb{o}\}\le r_k+\sup\{_k\Delta_{\epsilon_I}\}-\sup\{\mathbb{o}\}.$$ Now, let $$z=r_i+\sup\{_i\Delta_{\epsilon_I}\}-\sup\{\mathbb{o}\}.$$ Hence, $$0<z\le r_k+\sup\{_k\Delta_{\epsilon_I}\}-\sup\{\mathbb{o}\}.$$ Since $r_k<\sup\{\mathbb{o}\},$ we get $$0<z\le r_k+\sup\{_k\Delta_{\epsilon_I}\}-\sup\{\mathbb{o}\}<\sup\{\mathbb{o}\}+\sup\{_k\Delta_{\epsilon_I}\}-\sup\{\mathbb{o}\}=\sup\{_k\Delta_{\epsilon_I}\}.$$ 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 $$0<z<\sup\{_k\Delta_{\epsilon_I}\}$$ for any $k\ge i.$ This, in turn, means that $$0<zc^2<c^2\cdot \sup\{_k\Delta_{\epsilon_I}\}.$$ 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 $$\sup\{\mathbb{o}\}-r_k<zc^2.$$

We know that $r_{k+1}=r_k+c^2\cdot\sup\{_k\Delta_{\epsilon_I}\}.$ Consequently, $$\sup\{\mathbb{o}\}-r_{k+1}=\sup\{\mathbb{o}\}-r_k-c^2\cdot \sup\{_k\Delta_{\epsilon_I}\}<zc^2-c^2\cdot \sup\{_k\Delta_{\epsilon_I}\}<0.$$ Therefore, we have $$\sup\{\mathbb{o}\}-r_{k+1}<0.$$ Consequently, $$\sup\{\mathbb{o}\}<r_{k+1}.$$ 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 $$a=r_0<r_1<...<r_n<r_{n+1}=b.$$ Additionally, this partition also has the property that $$H_i(r_{i+1})-h_i(r_{i+1})<\epsilon_I$$ for any $0\le i<n+1.$ 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 $$L_a^b(f)<U_a^b(f).$$ By definition, $$u(f,\mathcal{R})\ge U_a^b(f),$$ $$l(f,\mathcal{R})\le L_a^b(f).$$ Consequently, $$l(f,\mathcal{R})\le L_a^b(f)<U_a^b(f)\le u(f,\mathcal{R}).$$ Rearranging yields $$0\le L_a^b(f)-l(f,\mathcal{R})<U_a^b(f)-l(f,\mathcal{R})\le u(f,\mathcal{R})-l(f,\mathcal{R}).$$ Therefore, $$0<U_a^b(f)-l(f,\mathcal{R})\le u(f,\mathcal{R})-l(f,\mathcal{R}).$$ This implies that there exists a $0<d<U_a^b(f)-l(f,\mathcal{R})$ 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. $\square$

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    $\begingroup$ Well, as soon as one $\delta$ works, so do all smaller positive numbers. $\endgroup$ – Randall Nov 8 at 15:47
  • $\begingroup$ 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? $\endgroup$ – TheGeometer Nov 8 at 15:50
  • $\begingroup$ I would be very interested in any proof which needs this specific set. Please edit your question to include this proof. $\endgroup$ – Carl Christian Nov 8 at 15:59
  • $\begingroup$ 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. $\endgroup$ – TheGeometer Nov 8 at 16:08
  • $\begingroup$ 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. $\endgroup$ – Omnomnomnom Nov 8 at 16:17

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