View source for Intermediate value theorem

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==Statement==
Let {{Top.|X|J}} be a ''{{link|connected|topology}}'' [[topological space]] and suppose {{M|f:X\rightarrow}}[[the real line|{{M|\mathbb{R} }}]] is a [[continuous function]], then{{rITTMJML}}:
* {{M|\forall x_1,x_2\in X\forall r\in\mathbb{R}\Big[\big(\text{Min}(f(x_1),f(x_2))\le r\le\text{Max}(f(x_1),f(x_2))\big)\implies\exists x\in X[f(x)\eq r]\Big]}}<ref group="Note">Recall that {{M|a\le b\le c}} really means {{M|(a\le b)\wedge(b\le c)}} - so {{M|a\le b}} ''and'' {{M|b\le c}}.</ref>
** In words: for any {{M|x_1,\ x_2\in X}} and any {{M|r\in\mathbb{R} }}, if {{M|r}} is between {{M|f(x_1)}} and {{M|f(x_2)}} then there exists an {{M|x\in X}} such that {{M|f(x)}} equals {{M|r}}.
==Proof==
* Let {{M|x_1,\ x_2\in X}} be given
** Let {{M|r\in\mathbb{R} }} be given
*** Define {{M|m:\eq\text{Min}(f(x_1),f(x_2))}} and let {{M|M:\eq\text{Max}(f(x_1),f(x_2))}}
***# Suppose that {{M|(m\le r)\wedge(r\le M)}} is ''false'', that it doesn't hold. Then by the nature of [[logical implication]] the implication evaluates to true regardless of the truth or falsity of the RHS
***# Suppose that {{M|(m\le r)\wedge(r\le M)}} holds, by the nature of logical implication we must show that in this case {{M|\exists x\in X[f(x)\eq r]}} for the implication to be true
***#* By [[the image of a connected topological space is connected]] we see that, as {{M|X}} is connected, so is {{M|f(X)\subseteq\mathbb{R} }}
***#** By [[a non-empty subset of the reals is connected if and only if it is a singleton or an interval]] we see that {{M|f(X)}} is an interval if we consider {{M|[a,a]}} as {{M|\{a\} }} - which is sensible.
***#*** As {{M|m\in f(X)}} and {{M|M\in f(X)}} we see that {{M|[m,M]\subseteq f(X)}} - note this may be a [[singleton]] or it may be an {{link|interval|reals}}
***#*** We can perform the rest of the analysis using [[case analysis]]
***#**** Notice first:
***#***** {{M|[m,M]\subseteq f(X)}} may be a [[singleton]], but it may not be
***#***** we are considering the case where {{M|r\in[m,M]}} this is okay because we know {{M|m\le r\le M}} - which is the very definition of {{M|r\in[m,M]}} - see [[closed interval]] for more. 
***#**** The case analysis follows, we have 2 main cases: {{M|[m,M]}} is a singleton and it is not, then some cases on {{M|r}}<ref group="Note" name="Case">We could get rid of one of our cases by noting that {{M|m\in[m,M]}} regardless of whether or not it is a singleton, we didn't do this to keep the "purity" of the case analysis</ref>
***#****# Suppose {{M|[m,M]}} is a singleton
***#****#* then as {{M|r\in[m,M]}} we must have {{M|r\eq m}} and {{M|r\eq M}} (as {{M|m\eq M}} for it to be a singleton)
***#****#** We see next that {{M|r\eq f(x_1)}} and {{M|r\eq f(x_2)}}
***#****#*** Choose {{M|x:\eq x_1}} ({{M|x_2}} would have worked) 
***#****#**** We see that {{M|f(x)\eq r}} as required for our choice of {{M|x}}
***#****# Suppose {{M|[m,M]}} is ''not'' a singleton. We create 3 cases: {{M|r\eq m}}, {{M|r\eq M}} and {{M|r\in(m,M)}}
***#****## {{M|r\eq m}}
***#****##* This means {{M|r\eq f(x_i)}} for some {{M|i\in\{1,2\} }}, as {{M|m:\eq\text{Min}(f(x_1),f(x_2))}} remember.
***#****##** Choose {{M|x:\eq x_i}}
***#****##*** We see that {{M|f(x)\eq f(x_i)\eq r}} so {{M|f(x)\eq r}} as required
***#****## {{M|r\eq M}}
***#****##* This means {{M|r\eq f(x_i)}} for some {{M|i\in\{1,2\} }}, as {{M|M:\eq\text{Max}(f(x_1),f(x_2))}} remember.
***#****##** Choose {{M|x:\eq x_i}}
***#****##*** We see that {{M|f(x)\eq f(x_i)\eq r}} so {{M|f(x)\eq r}} as required
***#****## {{M|r\in(m,M)}}
***#****##* As {{M|(m,M)\subseteq[m,M]\subseteq f(X)}} we see {{M|(m,M)\subseteq f(X)}}<ref group="Note">Really in this case we have {{M|(m,M)\subset[m,M]}} but {{M|\subseteq}} still works, remember to claim it's a [[proper subset]] means to claim they're not equal, we do not bother with this here, even though it is true</ref>
***#****##** so {{M|r\in f(X)}}
***#****##*** By definition of {{link|image|function}}, {{M|r\in f(X)\iff \exists p\in X[f(p)\eq r]}}
***#****##**** Choose {{M|x:\eq p}} that we know to exist from {{M|r\in f(X)\iff \exists p\in X[f(p)\eq r]}}
***#****##***** Now {{M|f(x)\eq r}} as required.
***#**** We see that in all cases (and their sub-cases) that: {{M|\exists x\in X[f(x)\eq r]}}, as required.
***#***** '''Note: ''' We could have tidied up the cases a bit, however we chose not to, as per<ref group="Note" name="Case"/>.
***#* We have seen that if {{M|(m\le r)\wedge(r\le M)}} that {{M|\exists x\in X[f(x)\eq r]}} indeed holds
*** The implication holds
** Since {{M|r\in\mathbb{R} }} was arbitrary the implication holds for all such {{M|r}}
* Since {{M|x_1,x_2\in X}} were arbitrary the implication holds for all such {{M|x_1,x_2}}
This completes the proof.
==Notes==
<references group="Note"/>
==References==
<references/>
{{Theorem Of|Topology|Functional Analysis|Analysis|Real Analysis|Metric Space}}