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Bishop states the constructive or approximate intermediate value theorem on page 40 of Constructive Analysis.

Approx. IVT: Let $f$ be a continuous map on an interval $I$ with $a,b \in I$ and $f(a) < f(b)$. Then for each $y \in [f(a),f(b)]$ and each $\epsilon > 0$, there exists $x \in [min\{a,b\},max\{a,b\}]$ such that $|f(x) - y| \leq \epsilon$.

The first line of the proof says "Since $f$ is continuous we must have $a \neq b$." Recall, in Bishop $a \neq b$ iff either $a < b$ or $b < a$. I'm having some trouble showing this first line rigourously although it seems obvious.

What I can show is that if $f$ is continuous and $f(a) < f(b)$ then $a = b$ leads to contradiction. Thus, $\lnot(a = b)$. But with out decidability of order on the reals I can not take the next step and conclude $a \neq b$. What am I missing?

ToucanIan
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  • I don't see what continuity has to do with this, but I would say that it is more of the fact that $f$ is a function. We note that the hypothesis of the statement is "Let $f$ be a continuous map on an interval $I$ with $a,b\in I$ and $f(a)<f(b)$".

    We wish to conclude that under these hypotheses that $a\neq b$. You can see this by the contrapositive that is if $a=b$, then $f(a)\not<f(b)$ (as $f(a)=f(b)$).

     – Steven Creech Jan 19 '22 at 04:53
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    @StevenCreech Somewhat confusingly, in Bishop's Constructive Analysis, $a \neq b$ does not denote the negation of $a = b$, but a different (and stronger - well, constructively stronger) relation. Your contrapositive only gives that if $f(a) < f(b)$, then "it is not the case that $a = b$" - which is a strictly weaker conclusion than $a \neq b$. – Z. A. K. Jan 19 '22 at 08:02
  • Thank you for making this point. – ToucanIan Jan 19 '22 at 16:34

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Let $\omega$ denote a modulus of continuity of $f$ on the interval $I$.

Since $f(a) < f(b)$, we have that $0 < f(b) - f(a)$, so in particular $\frac{|f(b) - f(a)|}{2} = \frac{f(b) - f(a)}{2} > 0$.

So set $\varepsilon = \omega\left(\frac{|f(b) - f(a)|}{2}\right)$. Since $f$ is continuous, we get that if $|b - a| < \varepsilon$ held, then we would have $|f(b) - f(a)| \leq \frac{|f(b) - f(a)|}{2}$, a contradiction. By Lemma 2.18, this means that $|b - a| \geq \varepsilon > 0$, and thus $b \neq a$.

Z. A. K.
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    Thank you for this insight. Sometimes these trivial results seem so difficult to prove because I feel like my intuition can’t be trusted. Something like lemma 2.18 seems invalid if I wasn’t sitting here staring at its proof. – ToucanIan Jan 19 '22 at 17:07
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    I understand your frustration @ToucanIan. Indeed, one of the motivations for Bishop's book was that it wasn't obvious (even to the top mathematicians of the time) that "everything works out"! But you will eventually learn the common tricks and develop good intuitions. Of course, it's much harder to learn than classical analysis: few people know any constructive analysis, while every math department has at least a few who know the classical theory very well. Moreover, the usual informal language of math has some classical bias in that constructive distinctions are difficult to make informally. – Z. A. K. Jan 19 '22 at 22:21
  • I’ve come up with another question. You conclude $b \neq a$ from $|b-a| \geq \epsilon > 0$. Of course this obvious but I wanted to prove it. As the absolute value is defined in terms of the max it suffices to prove: for any real $a,b$ and $c$ $max{a,b} \geq c$ implies $a \geq c$ or $b \geq c$. I’m having trouble with this proof. – ToucanIan Feb 04 '22 at 10:00
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    I've unfortunately not been able to prove the above claim (and I'm not sure how to proceed.) But I did realize that it is sufficient to prove $max{a,b} > c$ implies $a > c$ or $b > c$. This is straight forward. – ToucanIan Feb 17 '22 at 15:56
  • I've just noticed your comments. I'll try to take a look at your original question regarding the non-strict inequality tomorrow! – Z. A. K. Feb 17 '22 at 16:39