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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 ...
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 ...
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 ...
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 $3 \times B$, ...
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, ...
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 ...
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 ...
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 ...
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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