IDs 1-99

Evident.

Let $X$ be infinite and $A \subseteq X$ countably infinite. If $X$ has no limit points, $A$ is closed and for each $x \in A$, choose a nbd $V_x$ with $V_x \cap X = \emptyset$. Then $\{V_n\} \cup \{A^C\}$ is a cover. A finite subcover would imply $A$ is finite.

Contrapositively, suppose $\{V_n\}$ is a cover with no finite subcover. Then for $W_n = \bigcup_{m \le n} V_n$, each $X \setminus W_n$ is nonempty. So for each $n$, choose $x_n \in X \setminus W_n$. The sequence $\{x_n\}$ cannot have a converging subsequence: If $p \in X$, then $p \in V_m$ for some $m$ and so $x_n \notin V_m$ for all $n > m$.

For $f : X \to \R$ continuous, $\{f^{-1}(-n, n)\}$ is a countable cover.

Let $\{K_n\}$ be a countable compact cover of $X$ of which for all $x \in X$, some $K_n$ is a compact nbd of $x$. Let $K$ be compact. For each $x \in K$, let $n_x$ be the least element of which $K_n$ is a compact nbd. Then $\{\text{int}(K_{n_x})\}$ is a cover of $K$, which must have a finite subcover, and so finitely many $K_n$. Meaning $A \subseteq \bigcup_{n \le m} K_n$ for big enough $m$.

So define $K'_m = \bigcup_{n \le m} K_n$. Clearly $\{K'_m\}$ is a countable compact cover of $X$ and any compact is contained in some $K'_m$ by above.

Any closed set is compact, so any closure of a nbd is compact.

Take one nbd from the local basis. Its closure is compact.

Yeah.

Indeed.

Contrapositively, suppose $(x_n)$ is a sequence with $x_n \to p$ yet has infinitely many terms. If needed, take an injective subsequence. If $V$ is a nbd of $p$ that is $T_2$, then there’s a $n_0$ with $x_n \in V$ for $n \ge n_0$. So by another subsequence, we can assume $(x_n)$ is an injective converging sequence in a $T_2$ space.

For each $n$, it’s possible to construct a neighborhood $V_n$ which only contains $x_n$ and no other $x_m$, nor $p$: Let $x_n \in V$ and $p \in U$ with $U \cap V = \emptyset$. $x_m \in U$ for all $m \ge n_0$, so apply the Hausdorff condition finitely many times to terms $x_m < n_0$ to construct such $V_n \subseteq V$.

Now note that $x_{2n} \to p$ and $x_{2n+1} \to p$, so if $E = \bigcup V_{2n}$ and $O = \bigcup V_{2n+1}$, then any nbd of $p$ must intersect $E$ and $O$, so $p \in \overline{E} \cap \overline{O}$. Yet by definition, $E \cap O = \emptyset$, so the space is not extremally disconnected.

We just have to argue a $k$-netork is a network: Let $\mathcal{N}$ be a $k$-network. Singletons are compact. So in a $k$-network, for each $x$ in an open set $U$, we can find $\mathcal{N}^*_x \subseteq \mathcal{N}$ with $\{x\} \subseteq \bigcup \mathcal{N}^*_x \subseteq U$ (we don’t need $\mathcal{N}^*_x$ to be finite here, so no need for the axiom of choice, just take every set of $\mathcal{N}$ containing $x$). Thus,

$$U = \bigcup_{x \in U} \{x\} \subseteq \bigcup_{x \in U} \mathcal{N}^*_x \subseteq U \ \implies \ \text{$U$ is a union of sets in $\mathcal{N}$, so it is a network}$$

A subcover is a refinement. A finite subcover is star-finite.

If finitely many intersect a nbd around the point, finitely many will intersect the point.

If true for any covers, then true for countable ones.

If true for any covers, then true for countable ones.

A subcover is a refinement. A finite subcover is star-finite.

This is Paracompact $\implies$ Metacompact, just countable this time.

Let $A$ be an infinite set. Contrapositively, suppose every limit point of $A$ is not an $\omega$-accumulation point. If $p$ were to be a limit point of $A$, there exists a nbd $V$ of $p$ of which $V \cap A = \{ a_1, \dots, a_n \}$ is finite. But now for each $a_k$, there’s $V_k$, a nbd of $p$, so that $a_k \notin V_k$ (by being $T_1$). Then $W = V \cap V_1 \cap \cdots \cap V_n$ is a nbd of $p$ with $W \cap A = \emptyset$. So $A$ is an infinite set with no limit points.

By definition.

A $k$-netork is a network: This was done in (T11).

Let $f : X \to [0, 1]$ continuous with $f(A) = \{0\}$ and $f(b) = 1$. Then $U = f^{-1}([0, 1/2))$ and $V = f^{-1}((1/2, 1])$ are disjoint with $A \subseteq U$ and $b \in V$.

A space is a subspace of itself.

If it has a dispersion point… it has… a point.

Singletons are clopen.

The space is not a singleton.

This is obvious, so fun fact: The converse requires the continuum hypothesis.

Big brain stuff.

Singletons are compact.

The path between two distinct points has at least $\mathfrak c$ points.

The continuous map with $f(a) = 0$ and $f(b) = 1$ is not constant.

The discrete metric $d(x, y) = 1$ iff $x \ne y$ is complete: If \abs{x_n - x_m} < 1/2, then $x_n = x_m$.

Let $f : X \to [0, 1]$ with $f(a) = 0$ and $f(b) = 1$. Take $V_a = f^{-1}([0, 1/3))$ and $V_b = f^{-1}((2/3, 1])$. Then $f(\overline{V_a}) \subseteq \overline{f(V_a)} \subseteq [0, 1/3]$ and $f(\overline{V_b}) \subseteq \overline{f(V_b)} = [2/3, 1]$. This proves $\overline{V_a} \cap \overline{V_b} = \emptyset$.

If no disjoint nonempty closed sets exist, one of the disjoint closed sets is empty, and the other is contained in $X$, which is clopen.

Take a path between two points. It’s not constant.

Contrapositively, if all paths were constant, then any basis of path connected sets must be made up of singletons, proving it would be discrete.

Contrapositively, assume $A,B$ are disjoint, nonempty, and closed. Then we could separate them with $U,V$ open sets, disproving hyperconnectivity.

Let $X$ be partitioned into two sets with at least 2 elements each. One of them contains $p$, and so the other is contained in $X \setminus \{p\}$, so it is disconnected.

Let $\{N_n\}$ be the countable network and $A \subseteq X$. Ignore the network sets that don’t intersect $A$ and choose $x_n \in N_n \cap A$. Then $\{x_n\}$ has to be dense in $A$: For any $x \in A \cap V$, $x \in N_n \cap A \subseteq A \cap V$ for some $n$ so that $x_n \in A \cap V$.

If $f$ continuous, $f([0, 1/3])$ and $f([2/3, 1])$ are connected.

If any path connected component is open, then there can only be one (as they form a partition of $X$, disproving being connected). Take the path connected component $P$ of $x$. For $y \in P$, let $U$ be a connected nbd. Then any $z \in U$ is path connected to $x$ (just concatenate the paths $x \sim y$ and $y \sim z$). So $U \subseteq P$.

Every open set is dense, so $\overline{V} = X$ is trivially clopen for any open $V$.

Let $V$ be open. Then $\overline{V}$ is clopen, so it is either empty or $V$ is dense (no other nonempty open sets are disjoint with $V$).

$T_4$ is $T_2$ (hence, $T_1$) by definition.

Let $x \ne y$. $T_1$ implies $\{x\}$ and $\{y\}$ are closed, and normal implies there are $\{x\} \subseteq U$ and $\{y\} \subseteq V$ open sets with $U \cap V = \emptyset$. So it is $T_2$ and normal.