An easily verifiable proof of the Brouwer fixed point theorem

Theorem 1 (Brouwer [1], Hadamard [2]). Let n ∈ N and let g be a continuous mapping on [0, 1] n . Then there exists z ∈ [0, 1] n such that g(z) = z.

Theorem 1 is used in numerous fields of mathematics. So we can consider Theorem 1 is one of the most useful theorems in mathematics. There are many proofs of Theorem 1. For example, the proof based on the Sperner lemma [3] is very excellent in a geometric sense. See also Stuckless [4] and references therein.

On the other hand, in [4, page 4], it is said that “It has been estimated that while 95% of mathematicians can state Brouwer’s theorem, less than 10% know how to prove it.” This implies that Theorem 1 is very useful, however, the proofs we have obtained are not easy.

In this paper, motivated by this fact, we give a proof of Theorem 1. Our proof is so easy that most of the mathematicians can verify it. In our proof, we only use the Bolzano-Weierstrass theorem and the fact that an odd number plus an even number is odd. We do not need any geometric intuition. Therefore the authors believe that even undergraduate students are able to understand our proof.

Throughout this paper, we denote by N the set of all positive integers and by R the set of all real numbers. We define N(i, j) by

For x ∈ R n , (x) i denotes the i-th coordinate of x. For an arbitrary set B, we also denote by #B the cardinal number of B.

Let L be an arbitrary set, n ∈ N and k ∈ N(0, n). Let ℓ be a mapping from L into N(0, n). We call such a mapping a labeling. Then a subset B of L is called k-fully labeled if #B = k + 1 and ℓ(B) = N(0, k). The following lemma is very fundamental, however, it plays an important role.

Lemma 2. Let L be an arbitrary set, n ∈ N and k ∈ N(1, n). Let ℓ be a labeling from L into N(0, n). Let B be a subset of L with #B = k + 1 and ℓ(B) ⊂ N(0, k). Then the following hold: In the other case, where ℓ(b) = k, B is k-fully labeled and C is the only subset of B which is (k -1)-fully labeled. We have shown (i). From the above observation, (ii) is obvious.

In this section, we fix n, m ∈ N. We set n points e 1 , • • • , e n ∈ R n by

(1)

The following are obvious:

We sometimes write B = x 0 , • • • , x k . We note that B is a 0-string if and only if B = {0}. Theorem 3. Let ℓ be a Brouwer labeling from L n into N(0, n). Then there exists an n-string which is n-fully labeled.

We note that Theorem 3 is connected with the Sperner lemma [3]. Before proving Theorem 3, we need some preliminaries. Lemma 4. There exists exactly one 0-string which is 0-fully labeled.

Proof. Since ℓ(0) = 0, 0 is a 0-string which is 0-fully labeled.

Lemma 5. Let C be a (k -1)-string for some k ∈ N(1, n). Then there exists exactly one k-string which includes C.

Lemma 6. Let B be a k-string for some k ∈ N(1, n) and let C be a subset of B which is (k -1)-fully labeled. Then the following hold: Proof. Suppose B = x 0 , • • • , x k , x j = x 0 + j i=1 e σ(i) and B = C ⊔ {x h }. In the case of (i), it is obvious that h = k and σ(k) = k. So C is a (k -1)-string. By Lemma 5, only B is a k-string which includes C. In the case of (ii), we consider three cases:

In the first case, arguing by contradiction, we assume (

)-fully labeled. This is a contradiction. Thus (x k ) σ(1) < 1. Therefore

is another k-string and includes C. In the second case,

is another k-string and includes C. In the third case, arguing by contradiction, we assume

So C is not (k -1)-fully labeled. This is a contradiction. Thus (x 0 ) σ(k) > 0. Therefore

is another k-string and includes C. By (2) and (3), it is impossible that three k-strings include C.

By Lemmas 5 and 6, we obtain the following.

Lemma 7. Let C be a subset of L k which is (k -1)-fully labeled for some k ∈ N(1, n). Since #T 1 is odd, we obtain #S 1 is odd.

Then the following hold: (i) C is included by at most two k-strings. (ii) C is included by exactly one k-string if and only if C is a (k -1)-string.Lemma 8. Let k ∈ N(1, n). Assume there exist exactly odd (k -1)-strings which are (k -1)-fully labeled. Then there exist exactly odd k-strings which are k-fully labeled.Proof. 1 if and only if C is a (k -1)-string which is (k -1)-fully labeled. With double-counting, we count the number of (k -1)-fully labeled subsets in k-strings. Then by Lemmas 2 (i) and 7 (i), we have #S 1 + 2#S 2 = #T 1 + 2#T 2 .

Then the following hold: (i) C is included by at most two k-strings. (ii) C is included by exactly one k-string if and only if C is a (k -1)-string.Lemma 8. Let k ∈ N(1, n). Assume there exist exactly odd (k -1)-strings which are (k -1)-fully labeled. Then there exist exactly odd k-strings which are k-fully labeled.

Then the following hold: (i) C is included by at most two k-strings. (ii) C is included by exactly one k-string if and only if C is a (k -1)-string.