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Definition (Sequence)
A sequence of real numbers is a function $a:\mathbb{N}\to\mathbb{R}$.
- We'll often use the notation $a_n$ rather than $a(n)$ to denote the value of the sequence at $n$. We call $a_n$ the $n$th term of the sequence.
- We may refer to the sequence by writing $(a_n)$ or by writing $(a_1,a_2,\ldots)$.
- We may change the domain from $\mathbb{N}$ to $\mathbb{N}\cup \{0\}$ by adding in $0$, or we might remove several of the first entries from the sequence. When we do this, we'll use the notation $(a_n)_{n=m}^{\infty}$ where $m$ is the first term we want to consider.
Definition (Convergent Sequence)
Let $(a_1,a_2,\ldots)$ be a sequence of real numbers.
- We say that $(a_n)$ converges to $L$ and write $(a_n)\to L$ if and only if for every $\varepsilon>0$, there exists a real number $M$ such that for every $n\in \mathbb{N}$ we have $n> M$ implies $|a_n-L|<\varepsilon$.
- When $(a_n)$ converges to $L$, we call $L$ the limit of $(a_n)$.
- We say that $(a_n)$ is a convergent sequence if and only if $(a_n)$ converges to $L$ for some real number $L$.
- We say that $(a_n)$ is a divergent sequence if and only if $(a_n)$ is not a convergent sequence.
Problem 49: (Showing A Sequence Converges)
Consider the sequence $\ds (a_n)=\left(\frac{n+1}{n}\right)$. Prove that $(a_n)$ converges to $L=1$.
How do we determine if two sets have the same number of elements? If both sets are finite, the answer is simple as you can count the number of elements. However, what if both sets have infinitely many elements? The sets $\mathbb{Z}$ and $\mathbb{N}$ both have infinitely many elements. How do we compare the size of these sets?
- We could say there more than two times as many integers as there are natural numbers. This is because we can see an injection $f:\mathbb{N}\to\mathbb{Z}$ defined by $f(n)=n$, and this injection misses all the negative integers and zero.
To help us build some intuition, let's examine a classical fun way to look at this issue. It involves a rather strange hotel that has infinitely many rooms.
Problem: The Hilbert Hotel
Sally owns a rather strange hotel. This hotel has infinitely many rooms which are numbered 1, 2, 3, 4, .... Business has been good, and Sally manages to fill the entire hotel. Frank, Sally's brother, shows up at the main lobby and asks to get a room. The front clerk says, "Sorry, we're full." Luckily Sally overhears this comment and interrupts by saying, "Frank, I think we can find a room for you. We may have to ask some people to switch around rooms, but we can fit you in." Sally comes to you and asks you to find a room for Frank.
- Prepare new room assignments for every single occupant in Sally’s hotel so that after each occupant has moved to their new room, there will be an empty room available for Frank. In other words, if someone is currently in room $k$, which room do you want them to move to, and in which room will you put Frank. You’re allowed to move as many people in the hotel as you want, but you can’t make anyone share rooms.
- If George, who is Sally's cousin, had shown up at the same time as Frank, how could you have prepared the room assignments to make space for two empty rooms?
- If $n$ people show up to Sally's full hotel, how can you make room for $n$ extra people?
Definition: Cardinality
Suppose $A$ and $B$ are two sets. The cardinality of $A$ is written $|A|$.
- We say that the cardinality of $A$ is finite if $A$ has finitely many elements. Otherwise, we say that the cardinality of $A$ is infinite.
- If $A$ has finitely many elements, say $n$ elements, then we write $|A|=n$.
- If there is an injection $f:A\to B$, then we write $|A|\leq |B|$.
- If there is a surjection $g:A\to B$, then we write $|A|\geq |B|$.
- If there is a bijection $h:A\to B$, then we write $|A|=|B|$. Two sets have equal cardinalities if and only if there is a bijection between the two sets.
Problem: Several Countably Infinite Sets
What does the Hilbert Hotel Problem tell us about the cardinality of $\mathbb{N}$ and $\mathbb{N}\cup \{0\}$? What about the cardinality of $\mathbb{N}$ and $\mathbb{N}\cup \{0,-1,-2,-3,-4\}$? Which has a greater cardinality, or are they the same? Why?
Definition: Countably Infinite, Countable, Uncountable
Let $A$ be a set.
- We say that $A$ is countably infinite if $|A|=|\mathbb{N}|$.
- We say that $A$ is countable if either $A$ is a finite set or $A$ is countably infinite.
- We say that $A$ is uncountable if $A$ is not countable.
What happens when you start combining sets whose cardinality is infinite? Some might call the results below strange, while others might call them beautiful.
Problem: Finite Products of Countable Sets
Business has been really good for Sally, so her brother Frank decides to open another hotel across town, next to the river. Both Sally and Frank manage to fill their hotels completely every night (really good business). However, a week of excessive rain causes the river to flood, and Frank has to find lodging for all of his hotel guests (quite a feat). Frank calls Sally (whose hotel is full) and he tells her about the problem. Sally immediately says, ``Oh, I've got room for all your guests. Bring them over.'' Since you were the one who figured out how to make room for Frank, George, and $m$ friends, Sally figures you can find a way to fit in an extra infinite number of guests. She asks you to prepare the room assignments for all of Franks guests. Not wanting to disappoint her, you accept the challenge.
- Prepare new room assignments for every single occupant of both hotels. If someone is currently in room $k$ in your hotel, which room should they move to? If someone was in room $j$ of Frank's hotel, which room will you give them in your hotel? You can't make anyone share rooms. Sally would also like you to make sure the hotel is still full when you're done (if there are vacancies, it might look bad, we have to keep up appearances).
- George, remember Sally's cousin, decides to open a hotel as well (he puts it next to a lake). A month later, severe flooding causes both Frank and George to vacate their hotel. All three of Sally, Frank, and George have completely full hotels, but Sally tells the boys to bring all their guests over and she'll fit them all into the hotel. She then tells you to make room assignments.
Problem: Countable Products of Countable Sets
One night, Sally decides to take her hotel plan and sell the concept to potential investors. The idea is quite popular, and she fills each room in her hotel with someone who wants to build a hotel just like hers. For sake of ease, we'll name the new entrepreneurs $p_1, p_2, p_3, \ldots.$ Every entrepreneur goes out and opens a new hotel (businessman $p_k$ opens hotel $H_k$) that is just like Sally's. After a few months, every single hotel $H_k$ is filled to capacity. The business men $p_k$ decide to have a party in recognition of Sally, and as part of the party they decide that everyone should bring all their occupants to Sally's hotel on the same night (so there will be an infinite number of hotels, each bringing infinitely many guests). Sally decides to clear her hotel of all guests for the day, so her hotel is empty to start with. She asks you to prepare room assignments for all the inhabitants of all the hotels. Sally insists that you can do this.
- Your job is clear. When each guest arrives, they tell you which hotel $H_k$ they came from, and then they tell you which room number $j$ they were occupying at hotel $H_k$. How will you make the room assignments so that you can fill every single room in Sally's hotel with exactly one occupant? Which room will you assign guest $(j,k)$ to ($j$ was their old room assignment in hotel $H_k$)?
Problem: An uncountable set
One day, an alien space ship lands next to Sally's hotel. They had heard about her hotel and its amazing feats from an intergalactic newspaper. The alien ship is quite full, and Sally is having a hard time remembering names. She discovers that you can name all the aliens if you name them $a_x$ for each $x\in \mathbb{R}$. The aliens want to all stay in Sally's hotel, so Sally asks you to prepare room assignments for them all. You tell her it can't be done. She insists that it can be done, but you are firm in your convictions and tell her that it can't be done (you're not discriminating against aliens, rather you are just stating a fact). Sally doesn't believe you, so she prepares the room assignments herself. She brings you a list that says which alien should be in room 1, which alien to put in room two, etc. Before she takes it to the ship, she ask you to look over the list to make sure she hasn't missed anyone.
- Your job is clear. Show Sally why her room assignment plan is missing an alien. In other words, give her a counterexample that proves here list is incomplete.
When you complete this problem, you'll show that the cardinality of the real numbers is larger than the cardinality of the natural numbers. This proves that $\mathbb{R}$ is an uncountable set. There are different sizes of infinity.
Definition (Cardinality Of A Finite Set)
If $A$ is a finite set, then the cardinality of $A$, written $|A|$, is the number of distinct elements of $A$.
Problem 50: (Induction And Cardinality Of Cartesian Products)
If $A$ and $B$ are finite sets, then we have discussed in class why the cardinality of $A\times B$ is given by $$|A\times B|=|A|\cdot |B|.$$ Use induction to prove that for every $n\in \mathbb{N}$ if $A_1, A_2, \ldots, A_n$ are finite sets, then we have $$|A_1\times A_2\times\cdots\times A_n|=|A_1|\cdot|A_2|\cdots|A_n|.$$
Definition (Accumulation Point Of A Set)
Let $S$ be a set of real numbers. We say that a real number $a$ is an accumulation point of $S$ if for every $\varepsilon>0$, we have $N_\varepsilon^*(a)\cap S \neq \emptyset$.
The definition of accumulation point should look very similar to the definition of limit points. The next problem has you make this precise. You'll see a statement of the for "$P$ if and only if $Q$" in the problem below. This is shorthand for writing both "$P$ if $Q$" (so $Q\implies P$) and "$P$ only if $Q$" (so $P\implies Q$), which is the same as proving the two statements are logically equivalent.
Problem 51: (Accumulation Points Are The Same As Limit Points)
Let $S$ be a set. Prove that $a$ is an accumulation point of $S$ if and only if $a$ is a limit point of $S$. In other words, prove both of the following:
- If $a$ is an accumulation point of $S$, then $a$ is a limit point of $S$.
- If $a$ is a limit point of $S$, then $a$ is an accumulation point of $S$.
Let's now examine a problem related to limits of sequences. Some of you noticed in the definition of a limit of a sequence that we used the word "a" instead of "the." What we will prove now is that when a sequence converges, we can talk about "the" limit of the sequence instead of just "a" limit of the sequence.
Problem 52: (A Convergent Sequence Has A Unique Limit)
Let $(a_n)$ be a convergent sequence of real numbers. Prove that $(a_n)$ converges to a unique real number.
We now turn our attention to open sets. How does the union and intersection set operation behave when working with two open sets?
Problem 53: (Unions And Intersections Of Two Opens Sets Are Open)
Let $U_1$ and $U_2$ be open sets. Show that $U_1\cup U_2$ and $U_1\cap U_2$ are open sets.
Now that we've seen that unions and intersections of two open sets are open, what happens if we have more than two open sets? We need a formal definition of how to create a union and intersection of any collection of sets.
Definition (Unions And Intersections Of Arbitrarily Many Sets)
Suppose we have a large collection of sets. Each set has been given a name of the form $A_j$ where $j$ is an element of some nonempty set $J$. We call the collection $\mathrm{A} = \{A_j\mid j\in J\}$ an indexed family of sets with index set $J$. The union and intersection of all the sets in $\mathrm{A}$ are the sets given by $$\bigcup_{j\in J}A_j = \{x\mid x\in A_j \text{ for some } j\in J\}\text{ and } \bigcap_{j\in J}A_j = \{x\mid x\in A_j \text{ for every } j\in J\}.$$ When $J=\mathbb{N}$, then we'll often write the union and intersection using the notation $$\bigcup_{n\in \mathbb{N}}A_n = \bigcup_{n=1}^\infty A_n = A_1\cup A_2\cup A_3\cup\cdots\text{ and } \bigcap_{n\in \mathbb{N}}A_n = \bigcap_{n=1}^\infty A_n = A_1\cap A_2\cap A_3\cap\cdots.$$
Problem 54: (Unions And Intersections Of Nested Sets)
Let $n$ be a natural number and suppose that $A_1,A_2,\ldots, A_n$ are sets. Suppose also that $A_1\supseteq A_2\supseteq A_3\supseteq \cdots \supseteq A_n$.
- Prove that $\ds\bigcup_{i=1}^n A_i = A_1$.
- Make a conjecture that simplifies $\ds\bigcap_{i=1}^n A_i$. Then prove your conjecture.
Exercise (Induction With The Sum Of Cubes)
Prove that for every $n\in \mathbb{N}$, we have $$1^3+2^3+\cdots+n^3=\left(\frac{n(n+1)}{2}\right)^2.$$
Click to see a solution.
Consider the set $$S=\left\{n\in\mathbb{N}\mid 1^3+2^3+\cdots+n^3=\left(\frac{n(n+1)}{2}\right)^2\right\}.$$ We need to prove that $S=\mathbb{N}$. We proceed by induction. We know $1\in S$ because $1^3 =1$ and $\left(\frac{1(1+1)}{2}\right)^2 = (1)^2=1$. Assume for some $k\in \mathbb{N}$ that $k\in S$, which means we've assumed that $$1^3+2^3+\cdots+k^3=\left(\frac{k(k+1)}{2}\right)^2.$$ We need to prove that $k+1\in S$, which means we need to prove that $$1^3+2^3+\cdots+k^3+(k+1)^3=\left(\frac{(k+1)((k+1)+1)}{2}\right)^2.$$ We'll start with the expression $1^3+2^3+\cdots+k^3+(k+1)^3$ and modify it to obtain the right hand side of the equation above. We now compute $$\begin{align} 1^3+2^3+\cdots+k^3+(k+1)^3 &= \left(1^3+2^3+\cdots+k^3\right)+(k+1)^3 && (\text{group the first $k$ terms})\\ &= \left(\frac{k(k+1)}{2}\right)^2+(k+1)^3 && (\text{substitute using our assumption})\\ &= \frac{k^2(k+1)^2}{4}+\frac{4(k+1)^3}{4} && (\text{prepare to combine fractions})\\ &= \frac{(k+1)^2(k^2+4(k+1))}{4}&& (\text{combine and factor})\\ &= \frac{(k+1)^2(k^2+4k+4)}{4}&& (\text{expand})\\ &= \frac{(k+1)^2(k+2)^2}{4}&& (\text{factor})\\ &= \left(\frac{(k+1)((k+1)+1)}{2}\right)^2&& (\text{note $k+2=(k+1)+1$}) .\end{align}$$ This shows that if $k\in S$, then $k+1\in S$. By mathematical induction, we know that $S=\mathbb{N}$. This prove that for every $n\in \mathbb{N}$, we have $$1^3+2^3+\cdots+n^3=\left(\frac{n(n+1)}{2}\right)^2.$$
The formal principle of induction allows us to show that a set $S$ equals $\mathbb{N}$. This requires that we show $1\in S$ and if $k\in S$, then $k+1\in S$. What if the statement we wish to show is not true for every natural number, but is true for every natural number past some base case. So suppose we know that $4\in S$ and if $k\in S$ with $k\geq 4$, then $k+1\in S$. Can we conclude that $S$ contains every natural number greater than or equal to 4? If you examine the proof we used for induction, you'll find the quick answer to this question is "yes." So we now have a new form of induction.
- $b\in S$ and
- if $k\in S$ with $k\geq b$, then $k+1\in S$.
Problem 55: (Induction And $2^n\geq n^2$)
Use induction to prove that for every $n\in \mathbb{N}$ except for $n=3$ we have $2^n\geq n^2$.
When we have shown that something holds true for two objects, we can often use induction to prove the same fact holds true for any finite collection of those objects. The next problem has you do this.
Problem 56: (Unions And Intersections Of Finitely Many Opens Sets Are Open)
Use induction to prove for each $n\in \mathbb{N}$ if $U_1$, $U_2$, ..., $U_n$ are open sets, then $\ds \bigcup_{i=1}^n U_i$ is an open set.
Let's now return to studying functions.
Definition (Image $f(A)$ and Preimage $f^{-1}(B)$)
Consider the function $f:X\to Y$. Let $A$ be a subset of the domain $X$ and let $B$ be a subset of the codomain $Y$.
- The image of $A$ under $f$ is the subset of $Y$ defined by $$ \begin{align} f(A) &=\{y\in Y\mid y=f(a) \text{ for some }a\in A\}\\ &=\{f(a)\mid a\in A\} .\end{align}$$ This means that $y\in f(A)$ if and only if $y=f(a)$ for some $a\in A$.
- The preimage (or inverse image) of $B$ under $f$ is the subset of $X$ defined by $$ \begin{align} f^{-1}(B) &=\{x\in X\mid f(x)=b \text{ for some }b\in B\}\\ &=\{x\in X\mid f(x) \in B\} .\end{align}$$ This means that $x\in f^{-1}(B)$ if and only if $f(x)=b$ for some $b\in B$ if and only if $f(x)\in B$. Note that when the set $B$ contains a single element, then we write $f^{-1}(b)$ rather than $f^{-1}(\{b\})$.
We were just introduced to a new definition. We should try to apply this definition to a function we understand. The next problem has you do this.
Problem 57: (Function Notation With Sine)
Consider the function $f:\mathbb{R}\to\mathbb{R}$ defined by $f(x)=\sin x$.
- What are the sets $f(\mathbb{R})$ and $f([0,\pi/2))$?
- What are the sets $f^{-1}(0)$ and $f^{-1}([0,\pi/2))$?
- Find a set $A\subseteq \mathbb{R}$ so that $g:A\to \mathbb{R}$ defined by $g(x)=\sin(x)$ is injective.
- Find a set $B\subseteq \mathbb{R}$ so that $h:A\to B$ defined by $h(x)=\sin(x)$ is surjective.
The following theorem lists many facts that we could prove about the image and preimage of a function. We'll take some time to prove a few of them, but not all of them in class. You should be able to prove each of them for the final exam.
Theorem (Image And Preimage Properties)
Consider the function $f:X\to Y$.
- If $A\subseteq X$, then we have $A\subseteq f^{-1}(f(A))$.
- If $B\subseteq Y$, then we have $f(f^{-1}(B))\subseteq B$.
- If $A_1\subseteq X$ and $A_2\subseteq X$, then we have $f(A_1\cap A_2)\subseteq f(A_1)\cap f(A_2)$.
- If $A_1\subseteq X$ and $A_2\subseteq X$, then we have $f(A_1\cup A_2)=f(A_1)\cup f(A_2)$.
- If $B_1\subseteq Y$ and $B_2\subseteq Y$, then we have $f^{-1}(B_1\cap B_2)=f^{-1}(B_1)\cap f^{-1}(B_2)$.
- If $B_1\subseteq Y$ and $B_2\subseteq Y$, then we have $f^{-1}(B_1\cup B_2)=f^{-1}(B_1)\cup f^{-1}(B_2)$.
- We have $f(A)\subseteq B$ if and only if $A\subseteq f^{-1}(B)$.
- If $A_1\subseteq A_2\subseteq X$, then we have $f(A_1)\subseteq f(A_2)$.
- If $B_1\subseteq B_2\subseteq Y$, then we have $f^{-1}(B_1)\subseteq f^{-1}(B_2)$.
- If $B\subseteq Y$, then $f^{-1}(Y\setminus B)=X\setminus f^{-1}(B)$.
Problem 58: (Image And Preimage Properties 1 And 2)
Prove properties 1 and 2 for images and preimages. So prove for a function $f:X\to Y$ both of the following.
- If $A\subseteq X$, then $A\subseteq f^{-1}(f(A))$.
- If $B\subseteq Y$, then $f(f^{-1}(B)\subseteq B$.
Then give an example of a function $f:X\to Y$ and subsets $A\subseteq X$ and $B\subseteq Y$ where $A\neq f^{-1}(f(A))$ and $B\neq f(f^{-1}(B))$.
Problem 59: (Image And Preimage Property 3)
Prove property 3 for images. So let $f:X\to Y$. Then prove that
- If $A_1\subseteq X$ and $A_2\subseteq X$, then we have $f(A_1\cap A_2)\subseteq f(A_1)\cap f(A_2)$.
Finish by proving that if $f$ is injective, then equality holds.
Problem 60: (Image And Preimage Property 6)
Prove property 6 for preimages. So let $f:X\to Y$. Then prove that
- If $B_1\subseteq Y$ and $B_2\subseteq Y$, then we have $f^{-1}(B_1\cup B_2)=f^{-1}(B_1)\cup f^{-1}(B_2)$.
Problem 61: (Image And Preimage Property 7)
Prove property 7 for preimages. So let $f:X\to Y$. Then prove that
- We have $f(A)\subseteq B$ if and only if $A\subseteq f^{-1}(B)$.
Problem 62: (The Union And Intersection Of Infinitely Many Open Sets)
For each $n\in\mathbb{N}$, let $A_n = \left(0,1+\frac{1}{n}\right)$.
- What are the sets $\ds \bigcup_{n=1}^\infty A_n$ and $\ds \bigcap_{n=1}^\infty A_n$?
- Prove your claims.
Exercise (Unions And Intersections Of Opens Sets)
Prove each of the following:
- The union of any collection of open sets is open.
- The intersection of finitely many open sets is open.
Click to see a solution.
Let's first prove that the union of any collection of open sets is open. Let $J$ be a set and for each $j\in J$ let $U_j$ be an open set. This gives us a way to talk about an arbitrary collection of open sets. We must prove that $\ds\bigcup_{j\in J}U_j$ is an open set. So pick $\ds x \in \bigcup_{j\in J}U_j$. Since $x$ is an element of this union, we know that $x\in U_j$ for some $j\in J$. Since $U_j$ is an open set, we know we can pick $\varepsilon>0$ such that $N_\varepsilon (x)\subseteq U_j$. Since we know $U_j\subseteq \ds\bigcup_{j\in J}U_j$, this means $N_\varepsilon(x)\subseteq \ds\bigcup_{j\in J}U_j$. Since this entire argument holds for any $x\in \ds\bigcup_{j\in J}U_j$, we have shown that $\ds\bigcup_{j\in J}U_j$ is an open set.
We now prove that the intersection of finitely many open sets is open. One way to prove this is to refer to a previous problem where we used induction to prove this is true. Here is another proof. Let $n\in\mathbb{N}$ and suppose $U_1, U_2, \ldots, U_n$ are open sets. Let $x\in \ds\bigcap_{i=1}^n U_i$. To complete this proof, we must produce a postive $\varepsilon$ and prove that $N_{\varepsilon}(x)\subseteq \ds\bigcap_{i=1}^n U_i$. Pick $i\in\{1,2,\ldots,n\}$. Because $x\in \ds\bigcap_{j=1}^n U_j$, we know that $x\in U_i$. We assumed that $U_i$ is open, which means we can pick $\varepsilon_i$ such that $N_{\varepsilon_i}(x)\subseteq U_i$. Since the argument above holds for each relevant $i$, we pick $\varepsilon_i$ for each relevant $i$ so that $N_{\varepsilon_i}(x)\subseteq U_i$. Now comes the key part, namely we let $\varepsilon$ be the smallest of these positive values, which gives $$\varepsilon = \min\{\varepsilon_1, \varepsilon_2, \ldots, \varepsilon_n \}.$$ Clearly $\varepsilon>0$ by construction. In addition, because of how we defined $\varepsilon$, we know $N_{\varepsilon}(x)\subseteq N_{\varepsilon_i}(x)$ for each relevant $i$. This fact, together with the fact that $N_{\varepsilon_i}(x)\subseteq U_i$ for each relevant $i$, means we know $N_{\varepsilon}(x)\subseteq U_i$ for each relevant $i$. This fact proves that $N_{\varepsilon}(x)\subseteq \ds\bigcap_{i=1}^n U_i$, as needed. Our proof is complete (and should look very similar to the proof for two open sets).
Notice that the min function in the proof above can fail to produce a positive $\varepsilon$ if there is an infinite number of open sets. There is no guarantee that a minimum will even exist when a set has infinitely many elements. The problem before this exercise clearly shows us that the intersection of infinitely many sets does not have to be open, as we proved $\ds\bigcap_{n=1}^\infty \left(0,1+\frac{1}{n}\right) =(0,1] $.
Problem 63: (The Union And Intersection Of Infinitely Many Closed Sets)
For each $x$ such that $3<x<4$, let $A_x = [2,x]$.
- State an interval that equals each of $\ds \bigcup_{3<x<4} A_x$ and $\ds \bigcap_{3<x<4} A_x$.
- Prove your claims.
Definition (Function Composition)
Consider the functions $f:A\to B$ and $g:C\to D$. When $B\subseteq C$, then we know for each $a\in A$ that $f(a)\in B\subseteq C$. Since $f(a)\in C$, we can compute the quantity $g(f(a))$. If $B\subseteq C$, then we define the composition of $g$ and $f$ to be the new function $g\circ f : A\to D$ defined by $$(g\circ f)(a) = g(f(a)).$$
Problem 64: (The Composition Of Surjective Functions Is Surjective)
Let $A$, $B$, and $C$ be sets, and consider the functions $f:A\to B$ and $g:B\to C$. Prove that if both $f$ and $g$ are surjective, then $g\circ f$ is surjective.
Problem 65: (The Composition Of Injective Functions Is Injective)
Let $A$, $B$, and $C$ be sets, and consider the functions $f:A\to B$ and $g:B\to C$. Prove that if both $f$ and $g$ are injective, then $g\circ f$ is injective.
Problem 66: (Triangle Inequality)
For any real numbers $u$ and $v$, let $d(u,v)$ be the distance between $u$ and $v$, which means $d(u,v)=|u-v|=|v-u|$.
- Let $a,b,c\in\mathbb{R}$. Prove that $d(a,b)\leq d(a,c)+d(c,b)$.
- Let $x,y\in\mathbb{R}$. Use the previous result to prove that $|x+y|\leq |x|+|y|$.
For more problems, see AllProblems