Intermediate value theorem

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Statement

Let $(X,\mathcal{ J })$ be a connected topological space and suppose $f:X\rightarrow$$\mathbb{R}$ is a continuous function, then:

• $\forall x_1,x_2\in X\forall r\in\mathbb{R}\Big[\big(\text{Min}(f(x_1),f(x_2))\le r\le\text{Max}(f(x_1),f(x_2))\big)\implies\exists x\in X[f(x)\eq r]\Big]$
  • In words: for any $x_1,\ x_2\in X$ and any $r\in\mathbb{R}$, if $r$ is between $f(x_1)$ and $f(x_2)$ then there exists an $x\in X$ such that $f(x)$ equals $r$.

Proof

• Let $x_1,\ x_2\in X$ be given
  • Let $r\in\mathbb{R}$ be given
    • Define $m:\eq\text{Min}(f(x_1),f(x_2))$ and let $M:\eq\text{Max}(f(x_1),f(x_2))$
      1. Suppose that $(m\le r)\wedge(r\le M)$ 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
      2. Suppose that $(m\le r)\wedge(r\le M)$ holds, by the nature of logical implication we must show that in this case $\exists x\in X[f(x)\eq r]$ for the implication to be true
         • By the image of a connected topological space is connected we see that, as $X$ is connected, so is $f(X)\subseteq\mathbb{R}$
           • By a non-empty subset of the reals is connected if and only if it is a singleton or an interval we see that $f(X)$ is an interval if we consider $[a,a]$ as $\{a\}$ - which is sensible.
             • As $m\in f(X)$ and $M\in f(X)$ we see that $[m,M]\subseteq f(X)$ - note this may be a singleton or it may be an interval
             • We can perform the rest of the analysis using case analysis
               • Notice first:
                 • $[m,M]\subseteq f(X)$ may be a singleton, but it may not be
                 • we are considering the case where $r\in[m,M]$ this is okay because we know $m\le r\le M$ - which is the very definition of $r\in[m,M]$ - see closed interval for more.
               • The case analysis follows, we have 2 main cases: $[m,M]$ is a singleton and it is not, then some cases on $r$
                 1. Suppose $[m,M]$ is a singleton
                    • then as $r\in[m,M]$ we must have $r\eq m$ and $r\eq M$ (as $m\eq M$ for it to be a singleton)
                      • We see next that $r\eq f(x_1)$ and $r\eq f(x_2)$
                        • Choose $x:\eq x_1$ ($x_2$ would have worked)
                          • We see that $f(x)\eq r$ as required for our choice of $x$
                 2. Suppose $[m,M]$ is not a singleton. We create 3 cases: $r\eq m$, $r\eq M$ and $r\in(m,M)$
                    1. $r\eq m$
                       • This means $r\eq f(x_i)$ for some $i\in\{1,2\}$, as $m:\eq\text{Min}(f(x_1),f(x_2))$ remember.
                         • Choose $x:\eq x_i$
                           • We see that $f(x)\eq f(x_i)\eq r$ so $f(x)\eq r$ as required
                    2. $r\eq M$
                       • This means $r\eq f(x_i)$ for some $i\in\{1,2\}$, as $M:\eq\text{Max}(f(x_1),f(x_2))$ remember.
                         • Choose $x:\eq x_i$
                           • We see that $f(x)\eq f(x_i)\eq r$ so $f(x)\eq r$ as required
                    3. $r\in(m,M)$
                       • As $(m,M)\subseteq[m,M]\subseteq f(X)$ we see $(m,M)\subseteq f(X)$
                         • so $r\in f(X)$
                           • By definition of image, $r\in f(X)\iff \exists p\in X[f(p)\eq r]$
                             • Choose $x:\eq p$ that we know to exist from $r\in f(X)\iff \exists p\in X[f(p)\eq r]$
                               • Now $f(x)\eq r$ as required.
               • We see that in all cases (and their sub-cases) that: $\exists x\in X[f(x)\eq r]$, as required.
                 • Note: We could have tidied up the cases a bit, however we chose not to, as per.
         • We have seen that if $(m\le r)\wedge(r\le M)$ that $\exists x\in X[f(x)\eq r]$ indeed holds
    • The implication holds
  • Since $r\in\mathbb{R}$ was arbitrary the implication holds for all such $r$
• Since $x_1,x_2\in X$ were arbitrary the implication holds for all such $x_1,x_2$

This completes the proof.