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Topology Exam 1 — Practice Problems

Score: 6/50

Problem 1 (Score: 3/10)

$$ x \in A \subset U. $$

Prove that $\mathcal{A}$ is a basis for the topology on $X$.

Remark:
Do not simply state “This is a theorem in our book.” Provide a full proof.


Scratch work: A basis for a topology has 2 properties? Yeah. For each element in an open set, there is a basis that has that element and is completely contained in the open set.

For each intersection of arbitrary basis, there is another basis that is completely contained in the intersection.

Proof: Let X be a topological space. Let $\mathcal{A}$ be a collection of open sets in X such that each open set U in X and for each x in U, there is an element A of $\mathcal{A}$ such that x is in $A \subset U.$

Let i be an arbitrary element in an open set of X. Now, by our definition of $\mathcal{A}$, we have that $i \in A \subset U$ satisfying the first condition of Basis for a topology. For the intersection condition, we need that if there are two basis with an intersection, there is a basis completely contained in that intersection. Our construction of A limits intersections of basis to also be intersections of open sets. Because a topology is closed under intersections of open sets, there is an open set in the intersection and therefore there is a basis that is completely contained in the intersection. Hence, the second condition is satisfied.

Problem 2 (Score: 0/10)

Define what it means for a topological space $X$ to be Hausdorff. If $X$ is Hausdorff, then prove that a sequence in $X$ can converge to at most one point in $X$.

Hausdorff means that two distinct open sets have nonintersecting neighborhoods.

If we had convergence to two points say x1 and x2, I don’t know what convergence means here. LOL.

Problem 3 (Score: 0/10)

Let $X$ and $Y$ be topological spaces, and let $f : X \to Y$ be a function.
Suppose that for each closed set $C$ in $Y$, the preimage $f^{-1}(C)$ is closed in $X$.
Prove that $f$ is continuous.


Don’t know these definitions.

Problem 4 (Score: 1/10)

Let $X$ be a set and let $P(X)$ be its power set (the set of all subsets of $X$). Prove that there is no surjective function $f : X \to P(X)$.


Prove there is no surjection from a set to its power set. A surjective function is one that satisfies (for all y in Y, there is an x st f(x)=y) I think I get it, but this is blurry.

Proof: Let f be a surjective function from X to P(X).

I actually don’t know. If X is finite, it’s obvious because there are more elements in P(X) than X.

If X is not finite, okay, there is like a mapping that goes from X to P(X). Are we saying that it’ll never “catch up” to P(X)? I don’t know. This is so shit.

f is a function that goes from X to P(X) and is surjective.

Take the map from the an element x of the set. x maps to some subset of X. At this point, I think we have to do induction on the amount of x’s in the set. Let the cardinality of the set X be 1. Now, from X to P(X) is impossible as P(X) has cardinality 2 and X has 1.

Now, given a set of cardinality n doesn’t have a surjection from X to P(X), for n+1, the cardinality of P(X) increases by n and X by 1. Thus, by induction blah blah.

For infinite sets, I don’t know actually.

Problem 5 (Score: 2/10)

Prove that each of the following sets is countable.

(a) $\mathbb{Z}^+ \times \mathbb{Z}^+$

$f(\mathbb{Z}_+ \times \mathbb{Z}_+) = f(n \times m) 2^n3^m$

(b) $\mathbb{Q}^+ = \left\{ \frac{a}{b} \mid a,b \in \mathbb{Z}^+ \right\}$

we need the grid construction. I personally disagree with this proof.

(c) The set $J$ of all two-element subsets of $\mathbb{Z}^+$.

$f(a,b) = 2^a3^b$

Answers:

1

$$ x \in A \subset U. $$

Prove that $\mathcal{A}$ is a basis for the topology on $X$. This is in the book ctrl+f “The first condition for a basis is easy”

2

To prove cantor’s theorem:

Let D be the set of x in X such that x is not in f(x).

Now, we take a y in X such that f(y) = D and we find there is none.

Proving this is harder. Let’s see.

Let D be the constuction above, since f is surjection, there is a y in X with f(y) = D.

Now, if $y\not\in D$. $y \in f(y) = D$ If y in D, $y \not\in f(y) = D$

Things to learn:

Sameness of a topology. Hausdorff continuity