Suppose that $ f $ is continuously on $ (0.1) $ and $ f (0) = f (1) $. To let $ n $ be any natural number. Prove that there is a number $ x $ so that $ f (x) = f (x + frac {1} {n}). $

I was wondering if my proof is logically sound, especially the last piece. I tried to apply the logic of the intermediate value theorem, but I was curious to see whether the contradiction actually gave the desired result:

To let $ g $ be a function that $ g (x) = f (x + frac {1} {n}) – f (x) $. We want to show that there is one $ x in (0.1) $ so that $ g (x) = 0 $. We prove by contradiction.

Accept $ forall x in (0,1), g (x) <0. $ Then for $ i = 0, 1, 2, …, n $ we have that $ g ( frac {i} {n}) <0 $. Therefore, $ f (0)> f ( frac {1} {n})> f ( frac {2} {n})> …> f (1), $ which means, that $ f (0) neq f (1) $.

Following a similar logic follows from this $ g (x)> 0 $ can't keep for everyone $ x in (0.1) $.

Therefore it must exist $ a, b in (0.1) $ so that $ a $ and $ b $ have a different sign, i.e. $ g (a) leq 0 leq g (b) $. According to the intermediate value theorem exists $ c $ so that $ g (c) = 0 $and we're done.

Can I change the heading of the current chapter automatically?

In picture we can see that the previous chapter (title) in the header (I text in the cell, not automatically) and when start title 2, the header still titleIf I change title With title 2 Modify the entire document.

Intersects a circle at the points P, P & #; to the sides BC, CA and AB of the triangle ΔABC; Q, Q & # 39 ;; R, R & # 39 ;, and if AP, BQ, and CR are simultaneous, then AP & quot ;, BQ & quot; YCR & # 39; simultaneously.

Below is my code for problem 1.4 in CTCI. I would like a review of my code and if my approach to the problem is correct.

Problem:
Palindrome permutation: Write a function for a given string to see if it is a permutation
a palindrome. A palindrome is a word or phrase that is the same forwards and backwards. ON
Permutation is a rearrangement of letters. The palindrome does not have to be limited to just
Dictionary words.

def checkPalindromeAndPermutation(inputstr):
lengthOfInputString = len(inputstr)
counterforodd = 0
actualCharactersInInput = 0
inputstr = inputstr.lower()
hashTable = dict()
for i in inputstr:
if i != " ":
hashTable(i) = hashTable.get(i, 0) + 1
actualCharactersInInput = actualCharactersInInput + 1
print(hashTable)
for item in hashTable:
# if input has even length, but each character's frequency is not even, then it's not a plaindrome
if actualCharactersInInput % 2 == 0 and hashTable(item) % 2 != 0:
return False
# if input has odd length, but more than one character's frequency is greater than 1 , then it's not a plaindrome
if actualCharactersInInput % 2 == 1 and hashTable(item) % 2 == 1:
counterforodd = counterforodd + 1
if counterforodd > 1:
return False
return True
print("Answer : " , checkPalindromeAndPermutation("abc bac"))

To let $ B = Pi _ { alpha} B _ { alpha} $ is the direct product of finitely many $ A $ algebras $ f _ { alpha}: A to B _ { alpha} $, Then, when $ f: A to B $ is given by $ f (x) = (f _ { alpha} (x)) _ { alpha} $ show that $$
f ^ * (Spec (B)) = bigcup _ { alpha} f _ { alpha} ^ * (Spec (B))
$$

My proof: When all tensors are over $ A $ and for rent $ k $ be the rest field on a certain main ideal $ p in Spec (A) $ we have that

begin {align *}
f ^ {* – 1} (p) & = Spec (B otimes k) \
& = Spec left ( bigoplus _ { alpha} B _ { alpha} right) otimes k right) \
& = Spec left ( bigoplus _ { alpha} B _ { alpha} otimes k right) \
& = bigsqcup _ { alpha} Spec left (B _ { alpha} otimes k right) \
& = bigsqcup _ { alpha} f _ { alpha} ^ {* – 1} (p) quad quad (1)
end {align *}

So we have that $ f ^ {* – 1} (p) neq varnothing iff exists alpha , ,, , , f _ { alpha} ^ {* – 1} (p) neq varnothing $, That gives us that $$
f ^ * (Spec (B)) = bigcup _ { alpha} f _ { alpha} ^ * (Spec (B))
$$

My only concern is that there is a disjoint union in Equation (1). I was wondering if this disjointed union is more external than internal.