• #todo/school : Read §4.5; 📅 2024-11-13 ✅ 2024-11-13
• #todo/school : Do problems 4.5.1, 2, 3, 7;  4.6.5 📅 2024-11-13 ✅ 2024-11-14

4.5.1

Proof

See Intermediate Value Theorem#^a30f94.

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4.5.2

For all of these we may use that Continuous Image of Connected Set is Connected. We'll use the fact that all intervals are connected sets.

Proof

a. Possible. Use $f:(0,1)\to [0,1]$ defined as:

Here $f$ is continuous, and $f((0,1))=[0,1]$ as desired.

b. Impossible. Since $[a,b]$ is compact (since it's closed and bounded) then by Compactness Is Preserved By Continuity, then $f([a,b])$ is compact. But the open interval $(c,d)\equiv f([a,b])$ cannot be compact itself (open intervals don't have their endpoints).

c. Possible. Use $f:(0,1)\to [4,\mathrm{\infty })$ where $f(x)=\frac{-1}{x(x-1)}$. Clearly $f$ is continuous (it's a rational function $x\ne 0,1$) and one can show that it is bounded below by 4, and not above.

d. Impossible. Since $\mathbb{Q}$ is not connected, then by the contrapositive of Continuous Image of Connected Set is Connected then $\mathbb{R}$ must not be connected neither. However, it is.

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4.5.3

Proof

Let $[a,b]$ be where $a<b$. Since it satisfies the IVP, then $\mathrm{\forall }x<y\in [a,b]$ and $\mathrm{\forall }L\in (f(x),f(y))$ then $\mathrm{\exists }c\in (x,y)$ where $f(c)=L$, and further by being increasing then $f(x)\le f(y)$. Note we can apply $a,b$ to this immediately to get $f(a)\le f(b)$.

To show $f$ is continuous on $c\in [a,b]$, let $\epsilon >0$. Notice we'll want to choose some $\delta$ such that $0<|x-c|<\delta$ implies that $|f(x)-f(c)|<\epsilon$.

Notice that then there is some $L_{\ell }\in (f(a),f(c))\cap (f(c)-\epsilon ,f(c))$ (it's only empty when $f(a)=f(c)$, but then continuity is trivial. Otherwise by $f$ increasing then $f(a)<f(c)$ so then $(f(a),f(c))$ is non-empty). Thus since $a<c$ then $\mathrm{\exists }{c}_{\ell }\in (a,c)$ where $f(c_{\ell })=L_{\ell }$. We can similarly produce a $L_r\in (f(c),f(b))\cap (f(c),f(c)+\epsilon )$ such that since $c<b$ then $\mathrm{\exists }{c}_r\in (c,b)$ where $f(c_r)=L_r$.

Now notice that $|f(c)-f(c_{\ell })|<\epsilon$ by construction, and further $|f(c_r)-f(c)|<\epsilon$ in a similar manner. Now notice that since $f$ is increasing then $f(c_{\ell })\le f(c_r)$ so then $|f(c)-f(c_{\ell })|\le |f(c_r)-f(c)|$. If we let some $y\in (c_{\ell },c_r)$ then automatically $|f(c)-f(y)|<\epsilon$ if we use the delta-neighborhood $\delta =min\{c-c_{\ell },c_r-c\}$.

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4.5.7

Proof

Construct the nested intervals $I_n$ defined by:

Notice that since $f([0,1])\subseteq [0,1]$ then by induction then $I_1\supseteq I_2\supseteq \dots$ (so $I_{n+1}\subseteq I_n$). We have to show that each of these $I_n$'s is a closed interval first, but notice that since $f$ is continuous then since $[0,1]=I_1$ is connected then $f([0,1])=I_2$ is connected, and thus must be a closed interval (closed intervals are the only connected sets, mainly shown via an application of the Extreme Value Theorem).

By the Nested Interval Property then $\mathrm{\exists }x\in \bigcap _{n=1}^{\mathrm{\infty }}{I}_n\subseteq [0,1]$. As a result then $x=f(x)=f(f(x))\dots$ by construction, so then $x$ is a fixed point.

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4.6.5

Proof

Assume for contradiction it's not a jump discontinuity. We can only have a removable discontinuity, or an essential discontinuity.

If it was a removable discontinuity, then that would imply that $\underset{x\to c}{lim}f(x)=L$ exists, just $L\ne f(c)$. But the issue here is that using 45 IVT and Discontinuity Practice#4.5.3 then (because monotone implies increasing or decreasing) then we have to have $L=f(c)$ which implies continuity (which contradicts the assumption).

If instead it was an essential discontinuity, then at least one of the two sided limits don't exist. If that's the case then since the function is monotone (using what we proved yet again) then since one of the one-sided limits doesn't exist (goes to infinity) then the other limit must go to a point past infinity, which is a contradiction.

Proving it directly, we'll show that $\underset{x\to c^{-}}{lim}f(x),\underset{x\to c^{+}}{lim}f(x)$ both exist and are not equal to each other. For context, say $f:A\to \mathbb{R}$ where $A\subseteq \mathbb{R}$.

To show they exist, notice that since $f$ is monotone then $|f(x)-f(c)|>0$ always (for any $x\in A$). If we consider $S=\{|f(x)-f(c)|:x\in A\}$ then because the set is bounded below (by 0) then there is some infimum we denote $s=infS$. But since $S\subseteq A\subseteq \mathbb{R}$ then by R - the Reals and Completeness then $s\in S$, so then $\mathrm{\exists }{x}^{\prime }\in A$ where $s=|f(x^{\prime })-f(c)|$.

Now let $\epsilon >0$. Choose $\delta =min\{\epsilon ,s\}$. Then letting $x\in A$ where $0<|x-c|<\delta \le s$, we want to show that $|f(x)-f(c)|<\epsilon$. Notice that since $s=|f(x^{\prime })-f(c)|\ge |f(x)-f(c)|$ then in the case where $s<\epsilon$:

but when $s\ge \epsilon$ then $\delta =\epsilon \le s$ which is a contradiction to $s$ being an infimum (because if $\epsilon \ge s$ then $s$ is not the greatest lower bound, as now $\epsilon$ is a lower bound for $S$)

Showing what we showed is the opposite of a Functional Limit, so then we've shown the functional limit must exist for this one side. The proof for the other one-sided limit is very similar. Clearly we cannot have a removable discontinuity as the second paragraph describes, so we are done.

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