New answers tagged elementary-proofs
4
votes
Simpler proofs using the axiom of choice
The following theorem, relating path-connectedness and arc-connectedness (arc := injective path):
"Every path-connected Hausdorff space is arc-connected"
can be proven both with the axiom of ...
Community wiki
4
votes
Simpler proofs using the axiom of choice
For an orthomorphism $A$ in a (real or complex) Banach lattice $X$ the formula $\lvert A\rvert x=\lvert Ax\rvert$ for $x\ge0$ can be obtained relatively quickly if one uses the fact that $X$ is order ...
Community wiki
7
votes
Simpler proofs using the axiom of choice
Many examples from set theory are known, but here is a very basic (third-order) theorem from most ordinary mathematics:
"A regulated$f:[0,1]\rightarrow \mathbb{R}$ is bounded", (&)
where ...
Community wiki
21
votes
Simpler proofs using the axiom of choice
In Division by Three, Peter G. Doyle and John Horton Conway show, without invoking the Axiom of Choice, that for any sets $A$ and $B$, if there is a bijection between $3 \times A$ and $3 \times B$, ...
Community wiki
17
votes
Simpler proofs using the axiom of choice
Another classic example is the Schröder-Cantor-Bernstein theorem.
Theorem. If a set $A$ injects into $B$ and $B$ injects into $A$ then there is a bijection.
If AC holds, then this is nearly trivial, ...
Community wiki
14
votes
Simpler proofs using the axiom of choice
Here is a nice example.
Theorem. Space $\newcommand\R{\mathbb{R}}\R^3$ is the disjoint union of circles.
The AC proof is simply to enumerate all points in $\R^3$ in a well-ordered sequence in length ...
Community wiki
9
votes
Simpler proofs using the axiom of choice
There is an interesting class of problems that are provable in $ZF$ by "coincidence" in the sense that either (a fragment of) $AC$ holds and then it is provable from the proof in $ZFC$, or ...
Community wiki
7
votes
Simpler proofs using the axiom of choice
Axiom of choice is frequently used in Analysis through the Hahn-Banach theorem. An example is the existence of solution of Dirichlet's problem (=balayage problem). This can be proved without an ...
Community wiki
4
votes
Accepted
Analytic expression for the min value of $g(t):= \sqrt{(t-1)^2 + a^2}+ b|t|$ subject to $|t-1| \le c$
First, a few simplifications. Note that $g(-t)\ge g(t)$ and $|-t-1|\ge|t-1|$ if $t\ge0$. So, without loss of generality (wlog) $t\ge0$ and
$$g(t)=\sqrt{(t-1)^2 + a^2} + bt. \tag{1}\label{1}$$
Since $g(...
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