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From Ben:

I'd love to have each of you have something to present today. Here is who we'll have present.

  • 53.2 - Joey/Class - We did finish, and Shaughn said he'd type it up.
  • 56 - Emilee
  • 58.1 - Connor
  • 58.2 -
  • 59 -
  • 60 -
  • 61 -

The following students were absent

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.

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.

A similar proof will show that for each $n\in \mathbb{N}$ if $U_1$, $U_2$, ..., $U_n$ are open sets, then $\ds \bigcap_{i=1}^n U_i$ is an open set.
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 $f^{-1}(\{b\})$.
Theorem (Image And Preimage Properties)

Consider the function $f:X\to Y$.

  1. If $A\subseteq X$, then we have $A\subseteq f^{-1}(f(A))$.
  2. If $B\subseteq Y$, then we have $f(f^{-1}(B))\subseteq B$.
  3. If $A_1\subseteq X$ and $A_2\subseteq X$, then we have $f(A_1\cap A_2)\subseteq f(A_1)\cap f(A_2)$.
  4. If $A_1\subseteq X$ and $A_2\subseteq X$, then we have $f(A_1\cup A_2)=f(A_1)\cup f(A_2)$.
  5. If $B_1\subseteq Y$ and $B_2\subseteq Y$, then we have $f^{-1}(B_1\cap B_2)=f^{-1}(B_1)\cap f^{-1}(B_2)$.
  6. If $B_1\subseteq Y$ and $B_2\subseteq Y$, then we have $f^{-1}(B_1\cup B_2)=f^{-1}(B_1)\cup f^{-1}(B_2)$.
  7. We have $f(A)\subseteq B$ if and only if $A\subseteq f^{-1}(B)$.
  8. If $A_1\subseteq A_2\subseteq X$, then we have $f(A_1)\subseteq f(A_2)$.
  9. If $B_1\subseteq B_2\subseteq Y$, then we have $f^{-1}(B_1)\subseteq f^{-1}(B_2)$.
  10. 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.

  1. If $A\subseteq X$, then $A\subseteq f^{-1}(f(A))$.
  2. 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 $A_n = \left(0,1+\frac{1}{n}\right)$.

  1. What are the sets $\ds \bigcup_{n=1}^\infty A_n$ and $\ds \bigcap_{n=1}^\infty A_n$?
  2. Prove your claims.
Exercise (Unions And Intersections Of Opens Sets)

Prove each of the following:

  1. The union of any collection of open sets is open.
  2. 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 $A_x = [2,x]$.

  1. State an interval that equals each of $\ds \bigcup_{3<x<4} A_x$ and $\ds \bigcap_{3<x<4} A_x$.
  2. 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 $d(u,v)=|u-v|=|v-u|$.

  1. Let $a,b,c\in\mathbb{R}$. Prove that $d(a,b)\leq d(a,c)+d(c,b)$.
  2. Let $x,y\in\mathbb{R}$. Use the previous result to prove that $|x+y|\leq |x|+|y|$.
Both facts above we call the triangle inequality. Both facts basically state that the distance from point $A$ to point $B$ is less than or equal to the distance traveled if you take the shortest route from $A$ to $B$ that must also pass through a third point $C$. Equality holds if $C$ is already on the shortest path from $A$ to $B$, otherwise the distance must increase.

For more problems, see AllProblems