By definition.

By definition.

Any bijective function is a homeomorphism in discrete space. Just take the permutation that swaps $a$ and $b$.

Essentially the same proof as Fréchet Urysohn $\implies$ Sequential (T184). Just swap “sequential closure” with “radial closure”.

Fréchet Urysohn is stronger because it asserts there is a sequence, a.k.a a transfinite sequence of length $\omega$.

Any radially closed set must be, in particular, sequentially closed.

Any open set has more than one point.

Let $p$ be isolated and $\phi : X \to X$ with $\phi(p) = q$ a homeomorphism. Then $q$ is isolated. Do this for every $q$.

Suppose a set $A$ is sequentially closed. Let $p \in X$ and $(x_\alpha)_{\alpha < \lambda} \subseteq A$. Take $V$ a countable nbd of $p$. If $x_\alpha \to p$, then there is an ordinal $\mu$ for which $x_\alpha \in V$ for all $\alpha \ge \mu$. The set of ordinals $\mu \le \alpha < \lambda$ is isomorphic to some $\lambda'$. So we have $(y_\alpha)_{\alpha < \lambda'} \subseteq V$, still with $y_\alpha \to V$.

From a converging transfinite sequence within a countable set, we can extract a $\omega$-sequence that still converges to the same value, this was done in (T211). So we construct a sequence $(z_n)$ with $z_n \to p$, and so $p \in A$ as we wished to prove.

Given $p \in \overline{A}$, choose $D \subseteq A$ countable with $p \in \overline{D}$. Suppose $(x_\alpha)_{\alpha < \lambda}$ is some transfinite sequence in $D$ with $x_\alpha \to p$. We now construct a sequence in $D$ which converges to $p$ and we’re done (thanks to PatrickR).

If some $y \in D$ is cofinal in $(x_\alpha)$, this means any neighborhood $U$ of $x$ must contain $y$, so we can construct a constant sequence $y_n = y$ with $y_n \to p$. If no $y \in D$ is cofinal, then each $I_y = \{ \alpha < \lambda \ : \ x_\alpha = y \}$ is finite and $\lambda = \bigcup_{y \in D} I_y$ is a countable union of finite sets, so it is a countable ordinal. We can assume $\lambda$ to be a regular cardinal, so it is already a sequence.

If each local basis $\mathcal{V}_x$ is countable and $X$ is countable, then $\mathcal{B} = \bigcup_{x \in X} \mathcal{V}_x$ is a countable basis.

If $A$ and $B$ are disjoint closed sets, then $\cl{A} \cap \cl{B} = \emptyset$ and so $\{A, B\}$ is a discrete family.

Let $V$ be an order convex nbd of $p \in \cl{A}$. If $A = ({\leftarrow}, p)$ and $B = (p, {\rightarrow})$, then we could assume either $A \cap V$ has no maximum or $B \cap V$ has no minimum. Otherwise, then we could take $x = \max A$ and $y = \min B$ then $(x, y) \cap V = \{p\}$ would be a nbd of $p$ and so $p \in A$ (it would then suffice to take the constant sequence $x_\alpha = p$). Without loss of generality, say $A \cap V$ is infinite.

Using the axiom of choice, specifically Zorn’s lemma, we can take the collection of all well-ordered subsets of $A \cap V$, and any maximal element $W$ must be cofinal in $A \cap V$. Since this set is well-ordered, there is an isomorphism $f : \xi \to W$ from some ordinal $\xi$ and $x_\alpha = f(\alpha)$ is a sequence with $x_\alpha \to p$ by cofinality.

$\{x\}$ is a finite neighborhood of $x$.

For $x \ne y$, $\{x, y\}$ is discrete, so some nbd of $y$ does not contain $x$ and vice-versa.

If $K$ is infinite, let $A = \{x_n\} \subseteq K$ be a countable subset. But then $A$ is discrete and sets of the form $\{x_n\}$ form a cover for $A$ with no finite subcover, so $K$ can’t be compact.

Let $(x_n)$ be a sequence. By contradiction, suppose it has no converging subsequence and for each $k < \omega$, define $A[k] = \bigcup_{m \ge k} \cl{\{x_m\}}$. We first prove each $A[k]$ is closed. As the space is sequential, it suffices to take $(y_n)$ a sequence in $A[k]$ with $y_n \to y$ and prove $y \in A[k]$. Let $Y(m) = \{ n > m \ : \ y_n \in \cl{\{x_m\}}\}$ for each $m \ge k$. If some $Y(m) is infinite, we could construct a subsequence $(y_{f(n)})$ in $\cl{\{x_m\}}$ such that $y_{f(n)} \to y$, and so $y \in \cl{\{x_m\}} \subseteq A[k]$. Otherwise, each $Y(m)$ is finite and infinitely many are nonempty. So let $m_1$ be the least $m$ for which $Y(m) \ne \emptyset$ and $n_1 \in Y(m_1)$. Inductively, let $m_{j+1}$ be the least $m > n_j$ such that $Y(m) \ne \emptyset$ and $n_{j+1} \in Y(m_{j+1})$. Now we’ve chosen subsequences $(y_{n_j})$ and $(x_{m_j})$ such that $y_{n_j} \in \cl{\{x_{m_j}\}}$. For any $V$ nbd of $y$, there exists some $n_0$ such that $y_{n_j} \in V$ for $j \ge n_0$. But then $V$ is a nbd of $y_{n_j}$, so $x_{m_j} \in V$ for all $j \ge n_0$. As this holds for all $V$, that would imply $x_{n_j} \to y$, a contradiction!

So each $A[k]$ is closed, and as the space is countably compact, let $p$ be an accumulation point of $(x_n)$. Then clearly $p$ is a limit point of $A[k]$ for all $k$. Now consider the sets $X(n) = \{ m > n \ : \ p \in \cl{\{x_m\}} \}$. If $X(k) = \emptyset$ for some $k$, that would imply $p \notin A[k+1]$, which is impossible since $p \in \cl{A[k+1]}$ and $A[k+1]$ is closed. So let $m_1 = \min X(1)$ and $m_{k+1} = \min X(m_k)$. We’ve contructed a sequence $(x_{m_k})$ such that $p \in \cl{\{x_{m_k}\}}$ for all $k$. Then for any $V$ nbd of $p$, $x_{m_k} \in V$ and that would imply $x_{m_k} \to p$.

Source: Sequential space methods, Kremsater

Take $x \in W \subseteq K$ with $V$ open and $K$ compact. Let $U$ and $V$ be a separation of $x$ and the closed set $W^C$. In particular, $\cl{U} \subseteq K$. The sets of the form $U \cap O$ where $O$ is open forms a local basis for $x$ and each closure $\cl{U \cap O} \subseteq K$ is compact.

Contrapositively, suppose there is no nbd of $x$ which doesn’t contain $y$ as well. Then for the constant sequence $x_n \coloneqq x$, $x_n \to x$ and $x_n \to y$.

If $f : K \to X$ continuous and $K$ compact, $f(K)$ is compact, so it is closed by KC.

$T_2$ implies $\Delta \subseteq X^2$ is closed. So $\Delta \cap K$ is trivially closed in $K$ for any $K \subseteq X^2$.

Let $\{U_n\}$ and $\{V_n\}$ be countable local bases of $x \ne y$. Contrapositively, suppose $U_n \cap V_n \ne \emptyset$ for all $n$, and choose $z_n \in U_n \cap V_n$. Then $z_n \to x$ and $z_n \to y$.

$\R$ is homeomorphic to $\R^1$.

If $X \simeq Y \subseteq \R$, for each $x \in X$, we shall assume $x$ is not minimum or maximum element of $L$, so that it has a local basis of sets of the form $(a, b) \cap X$ for $a,b \in L$ (otherwise the proof is similar but the intervals are either $[x, b)$ or $(a, x]$). Then there must be some $c \in (a, x)$ with $c \notin X$, otherwise $(a, x) \subseteq X$ would be a path connected subset¹. Similarly, there must be some $d \in (x, b) \setminus X$ so that $x \in (c, d) \cap X = [c, d] \cap X$ is a clopen nbd of $x$ in $X$. So the clopen nbds of $X$ form a basis.

(1): This is the one time in the proof where we use the fact that we’re in $\R$, because connected $\iff$ path connected in $\R$. Otherwise, we could construct a very similar proof for (Totally disconnected ∧ GO-space) $\implies$ Zero dimensional

Not sure why they expected locally here. If $f : X \to f(X) \subseteq \R$ is a homeomorphism and $X$ connected, then $f(X)$ is connected. Every connected set in $\R$ is path connected, so $f(X) \simeq X$ is path connected.

Compact sets are countably compact.

Suppose $A$ is countably compact yet not closed. $X$ is sequential, so there is a sequence $(x_n)$ in $A$ and $p \in \cl{A} \setminus A$ with $x_n \to p$. Then $(x_n)$ has some accumulation point $y \in A$. There cannot be infinitely many $n$ for which $x_n = y$, otherwise we could extract a subsequence with $x_{n_k} \to y$, and by US, that would imply $y = p$. Therefore, by taking a subsequence if necessary, we can assume $y \ne x_n$ for all $n$. But then $y \notin B = \{x_n \ : \ n < \omega \} \cup \{p\}$ yet $y \in \cl{B}$. So $B$ isn’t closed, which would imply there is some sequence $(y_n)$ in $B$ with $y_n \to q$ and $q \in \cl{B} \setminus B$. Just as before, there can only be finitely many $n$ with $y_n = x_m$ or $y_n = p$ for any $m$. So we could extract a subsequence $(y_{n_k})$ such that $y_{n_k} = x_{m_k}$ is also a subsequence of $(x_n)$ and so $y_{n_k} \to p = q$, a contradiction.

Fix $p \in X$. If $p$ is isolated, $\{\{p\}\}$ is a local basis. Otherwise, $p \in \cl{X \setminus \{p\}}$ and there is a countable $D \subseteq X \setminus \{p\}$ with $p \in \cl{D}$. So consider the intervals of the form $[p, x)$, $(x, p]$, or $(x, y)$ for $x,y \in D$ which are open in $X$. This must be countable and it is a local basis of $p$ since $p \in \cl{D}$ and it isn’t isolated in $X$.

By definition: $\sigma$-compact $\iff$ countable union of compacts $\{K_n\}$. Hemicompact $\iff$ countable union of compacts $\{K_n\}$ such that any compact set lies inside some $K_m$.

Globally implies locally.

This is just “Has a countable $k$-network $\implies$ Has a countable network” (T11) with $T_3$ added.

This is just “Has a countable network $\implies$ Has a $\sigma$-locally finite network” (T29) with $T_3$ added.

This is just “Has a countable $k$-network $\implies$ Has a $\sigma$-locally finite $k$-network” (T352) with $T_3$ added.

This is just “Has a $\sigma$-locally finite $k$-network $\implies$ Has a $\sigma$-locally finite network” (T34) with $T_3$ added.

Just take one element from the local basis.

The proof is literally the same as (T224). I’d like to express my discontent with $\pi$-base’s usage of the term “local basis”. To me, a local basis is by definition a collection of open sets, but what they mean by “locally compact” is: Each $x$ has a collection of compact sets whose interiors form a local basis for $x$.

$\mathcal{P}(X) = \{\emptyset, X\}$ if and only if it has 0 or 1 point.

There’s only one possible topology.

There’s only one possible topology.

If you have an infinite amount of apples, then you have at least 2 apples.

Any open cover must contain $X$ as an element.

Define $d(x, y) = 1$ if $x,y$ belong to the same set in the partition, and 0 otherwise. This is a pseudometric.

Some point has a neighborhood not containing another point.

A space is a subspace of itself.

It suffices to prove any $F_\sigma$ set is Lindelöf, as that implies every open set is Lindelöf in a $G_\delta$ space. This follows from two basic topological facts: (1) Closed sets of Lindelöf spaces are Lindelöf: If $A \subseteq X$ is closed and $\mathcal{U}$ is a cover of $A$, then $\mathcal{U} \cup \{A^C\}$ is a cover of $X$ with a countable subcover, which induces a countable subcover of $A$ (2) A countable union of Lindelöf spaces $\{L_n\}$ is Lindelöf: If $\mathcal{U}$ is a cover of $L = \bigcup_{n < \omega} L_n$, then it is a cover of each $L_n$, which gives us a countable subcover $\mathcal{U}_n$ of $L_n$, and so $\bigcup_{n < \omega} \mathcal{U}_n$ is a countable subcover of $L$.

Let $A$ be closed. Then there is a continuous map $f : X \to [0, 1]$ with $f^{-1}(\{0\}) = A$. Thus, each $V_n = f^{-1}([0, 1/n))$ is open and $A = \bigcap_{n=1}^\infty V_n$ is a $G_\delta$ set.

We prove any closed set $A$ is a zero set. Let $A = \bigcap_{n=1}^\infty V_n$ be a $G_\delta$ set. Without loss of generality, assume $\{V_n\}$ is decreasing. Define $A_n = V_n^C$, and by normality, let $f_n : X \to [0, 1]$ a function where $f_n(A) = \{0\}$ and $f_n(A_n) = \{1\}$. We define $f : X \to [0, 1]$ as

$$f(x) = \sum_{n=1}^\infty \frac{f_n(x)}{2^n}$$

It already has the property $f^{-1}(\{0\}) = A$ as we wished, since $f(x) = 0$ iff $f_n(x)$ for all $n$ and $x \notin A$ implies $x \in A_n$ for some $n$ and $f_n(x) \ne 0$. To prove continuity, we use the Weierstrass M-test. Each $g_n(x) = f_n(x)/2^n \le 2^{-n}$ and $\sum 2^{-n} = 1$ converges, so $\sum g_n(x)$ converges absolutely and uniformly. As each finite sum $\sum_{n=1}^N g_n(x)$ is continuous, their uniform limit $f$ is continuous.

We shall show $X$ is normal and $G_\delta$ (T257). The former is a result of (Regular ∧ Lindelöf) $\implies$ Normal, and the proof of this fact can be seen in the second-half of (T26). There, we merely use the fact that the space is regular and closed sets are Lindelöf, which is clearly true in a Lindelöf space.

Now let $A$ be closed. By regularity, for each $x \notin A$ there is a pair $x \in U$ and $A \subseteq V$ of open sets with $U \cap V = \emptyset$. By joining all such pairs $(U, V)$ into a collection $\mathcal{W}$, we see that $\pi_1(\mathcal{W})$ is a cover of $A^C$. By hereditarily Lindelöf, there is a countable subcollection $\{(U_n, V_n)\}$ in $\mathcal{W}$ which, for any $x \notin A$, there is some $n$ for which $n \notin V_n$. Conversely, $x \in A$ implies $x \notin V_n$ for all $n$. So $A = \bigcap_{n < \omega} V_n$ is a $G_\delta$ set.

Singletons form a network.

Let $\{N_k\}$ be the countable network. Let $A \subseteq X$ and $\mathcal{U}$ be an open cover of $A$. We know for any $x \in U \in \mathcal{U}$ there is a $k$ such that $x \in N_k \subseteq U$. Let $I$ be the set of $k < \omega$ such that $N_k \subseteq U$ for some $U \in \mathcal{U}$, and for each $k \in I$, choose one such $U_k$. Then $\{U_k\}_{k \in I}$ is a countable subcover.

In $R_0$ spaces, points $x,y$ are indistinguishable iff $\cl{\{x\}} = \cl{\{y\}}$. Thus, if $x \in V$ and $V$ is open, $\cl{\{x\}} \subseteq V$. So $V = \bigcup_{x \in X} \cl{\{x\}}$ is an $F_\sigma$ set, as $X$ is countable.

If no nonempty disjoint open sets exist, $R_1$ implies all points are indistinguishable.

Let $A$ be the set of points which have a countable nbd. Then $A$ is locally countable and Lindelöf, so it is countable (T499). This must imply $B = A^C$ is empty, as otherwise it would need to have an isolated point $p \in B$ which has no countable nbd, so $p \in U$ and $U \cap B = \{p\}$ would imply $U \setminus \{p\} \subseteq A$ is uncountable (thanks to Almanzoris, I had proved it on my own but it was far more complicated than this nice argument).

Every metric is a pseudometric.

By contraposition, if $d(x, y) = 0$, then they both have the same neighborhoods.

Globally implies locally.

Let $\mathcal{V}$ be all nbds of $x$. Then $\bigcap \mathcal{V}$ is an open set. For any $y \ne x$, there is some nbd $U_y$ of $x$ with $y \notin U_y$, so $\{x\} = \bigcap \mathcal{V}$.

We prove it is normal and $G_\delta$ (T257). For the latter, let $A$ be any closed set, and define $V_n = \bigcup_{x \in A} B(x, 1/n)$. Let $p \in \bigcap_{n=1}^\infty V_n$. For each $n$ there is an $x_n \in A$ such that $p \in B(x_n, 1/n)$ and therefore, $x_n \to p$. Since $A$ is closed, $p \in A$. This proves $A = \bigcap_{n=1}^\infty V_n$ is $G_\delta$.

To prove normality, we use the fact that for a closed set $Y$, $d(Y, x) = 0 \implies x \in Y$. For any disjoint closed sets $A$ and $B$, define $r_x = d(B, x)/2$ for each $x \in A$ and $s_y = d(A, y)/2$ for each $y \in B$. Define $U = \bigcup_{x \in A} B(x, r_x)$ and $V = \bigcup_{y \in B} B(y, s_y)$. It’s clear $A \subseteq U$ and $B \subseteq V$. If $p \in U \cap V$, then $d(x, p) < r_x$ and $d(y, p) < s_y$ for some $x \in A$ and $y \in B$. But then $d(x, y) < r_x + s_y = d(B, x)/2 + d(A, y)/2 \le d(x, y)$ is a contradiction.

A countable basis is a local basis for every point.

Any network of open sets is a $k$-network. If $K \subseteq \bigcup \mathcal{N}^* \subseteq V$ for some $K$ compact, we can take a finite subcollection of $\mathcal{N}^*$. Any basis is a network. Therefore, a countable basis is a countable $k$-network.

Let $\mathcal{U}$ be a $k$-cover and $\mathcal{B}$ a countable basis. For any compact $K$, $K \subseteq U$ for some $U \in \mathcal{U}$, which can be written as a union of basis elements $U = \bigcup \mathcal{B}^*$. This is an open cover of $K$, so we can extract a finite subcover. This allows to conclude that if for each finite selection of basis elements, we select one element $U$ which contains their union, then this must be a $k$-subcover, and the collection of all finite selections of basis elements $\bigcup_{n < \omega} \mathcal{B}^n$ is countable.

We use the axiom of choice (specifically the well-ordering principle) to denote a bijection $f : X \to \alpha$ where $\alpha$ is some initial ordinal. If $\alpha < \omega$, the order topology on $\alpha$ is discrete, so $f$ induces an order in $X$ which is discrete. If $\kappa = \card{\alpha}$ is infinite, then there is a bijection $g : \alpha \to \alpha \times \Z$ (the well-ordering of $\alpha$ allows us to construct a bijection from $\alpha \times \Z$ to $\alpha$ and then apply Schröder–Bernstein). The dictionary order on $\alpha \times \Z$ is discrete since the element $(\beta, n)$ is between $(\beta, n-1)$ and $(\beta, n+1)$, so the bijection $g \circ f$ yields a discrete order in $X$.

Let $A$ be countable and dense. Let $\mathcal{B}$ be the finite intersections of open rays of the form $(x, {\rightarrow})$ or $({\leftarrow}, x)$ for $x \in A$. For any $z \in X$, assume it is not a minimum or maximum of $X$ and suppose $z \in (a, b)$. The proof is similar for those cases by taking $[z, b)$ or $(a, z]$. Then $(a, z) \ne \emptyset$, as otherwise $X = ({\leftarrow}, z) \cup (a, {\rightarrow})$ would be disconnected. Furthermore, there must be some $x \in A$ in this interval, otherwise some $w \in (a, z)$ with $(a, z) \cap A = \emptyset$ would imply $w \notin \cl{A}$. Similarly, let $y \in (z, b)$ and so $z \in (x, y) \subseteq (a, b)$. Since $(x, y) \in \mathcal{B}$, this is a countable basis.

If $A \subseteq X$ is discrete, it can be covered by nbds which contain one element each. So if $A$ were to be Lindelöf, it has to have countably many elements.

Let $\{K'_n\}$ be the compact sets of which every compact $K \subseteq K_n$ for some $n$, and define $K_n = \bigcup_{n \le k} K'_k$, which has the same property. Let $\{U_n\}$ be a local basis for some $x$ and $V_n = \bigcap_{k \le n} U_k$ is also a countable local basis. We wish to prove $x \in V_n \subseteq K_n$ for some $n$. By contradiction, assume there is no such $n$. That means we can extract a sequence $x_n \in V_n \setminus K_n$. Then $x_n \to x$ and so $K = \{x_n \ : \ n < \omega\} \cup \{x\}$ is compact. But then $K \subseteq K_n$ for some $n$, which contradicts $x_n \notin K_n$.

$T_2$ is $T_0$ with $R_1$.

$x$ and $y$ are indistinguishable iff either $x \notin \cl{\{y\}}$ or $y \notin \cl{\{x\}}$. Suppose the former. Then regularity gives us $x \in U$ and $\cl{\{y\}} \subseteq V$ with $U \cap V = \emptyset$, which is clearly a separation of $x$ and $y$.

Any two distinct points are distinguishable in a $T_0$ space.

If $\mathcal{V}$ is the collection of all nbds of $x$, $W = \bigcap \mathcal{V}$ is the smallest nbd of $x$, so $\{W\}$ is a compact local basis of $x$: for any $\mathcal{W}$ open cover of $W$, there must be some $V \in \mathcal{W}$ with $x \in V$, and then $\{V\}$ is a subcover.

If $\mathcal{V}$ is the collection of all nbds of $x$, $W = \bigcap \mathcal{V}$ is the smallest nbd of $x$, so $\{W\}$ is a local basis of $x$.

Immediate by definition.

By definition, $T_1$ is $T_0$ and $R_0$.

$T_0$ ensures any two points are distinguishable. This is an if and only if.

To prove $R_0$, it suffices to prove $x \notin \cl{\{y\}} \implies y \notin \cl{\{x\}}$. Note that $\cl{\{y\}} = \bigcap_{n < \omega} V_n$ is a $G_\delta$ set, so there exists some $n$ such that $x \notin V_n$, yet $y \in V_n$. That proves $y \notin \cl{\{x\}}$.

A countable union of subsets in a finite space can be assumed to be a finite union by removing duplicates. Now we just use the fact that for finitely many closed sets $\{A_1, \dots, A_n\}$, that $\text{int}(\bigcap_{k=1}^n A_i) = \bigcap_{k=1}^n \text{int}(A_i)$. That is, a finite union of nowhere dense sets is nowhere dense.

If $X = \{x_1, x_2, \dots\}$ is countable, define $K_n = \{ x_1, x_2, \dots, x_n \}$. Then $X = \bigcup_{n=1}^\infty K_n$ and every compact/finite set is contained in some $K_n$.

We first show the space is Alexandrov. Note that if $\{V_i\}$ is any family of open sets, and $K$ is any compact subspace, $K \cap V_i$ is open in $K$ and also finite. Thus, by removing duplicates, the family $\{K \cap V_i\}$ is finite and so $\bigcap_{i} (K \cap V_i)$ is a finite intersection of open sets. Therefore, $K \cap \bigcap_{i} V_i$ is open for every compact $K$ and so by the $k_1$-space property, $\bigcap_{i} V_i$ is open. Being Alexandrov, let $V$ be the smallest nbd of $x$. Then it must be compact (T284), which is finite.

Cover a compact set $K$ with finite nbds. It has a finite subcover of finite sets, so it must be finite.

Important result to solve the Riemann hypothesis.

Indiscrete spaces are connected and being indiscrete is hereditary.

Every point $p$ is associated with the constant map $x \mapsto p$, which is continuous. So $X$ has at most as many continuous self-maps.

It has finitely many self-maps (continuous or not).