Contrapositively, let $f : X \to \R$ is continuous and non-constant. So $f(x) = \alpha$ and $f(y) = \beta$ with $\alpha \ne \beta$. Clearly $A_1 = f^{-1}(\alpha)$ and $A_2 = f^{-1}(\beta)$ are disjoint. They are zero sets, since $A_1 = g^{-1}(0)$ for $g(x) = f(x) - \alpha$ and similarly, $A_2$ is a zero set. So $A_1 \subseteq B_1$ and $A_2 \subseteq B_2$ with $B_1,B_2$ disjoint clopen sets, by strongly zero-dimensional. In particular, $x \in B_1 \subset X$, so the space is not connected.

If you remove “countable” from the definition of pseudonormal, that’s precisely one property of normal spaces. If $H$ is closed and $H \subseteq U$ is open, then $H \cap U^C = \emptyset$, so there are open sets $H \subseteq V$ and $U^C \subseteq V'$ with $V \cap V' = \emptyset$. In particular, $\cl{V} \subseteq U$.

Let $x \notin A$ with $A$ closed. Being $T_1$, $\{x\}$ is a countable closed set, and $\{x\} \subseteq U = A^C$ is open. So let $V$ be open with $x \in V \subseteq \cl{V} \subseteq A^C$. Then $A \subseteq \cl{V}^C = W$ with $V \cap W = \emptyset$.

Let $H = \{x_n\}$ and $H \subseteq U$ open. Then $x_n \notin F = U^C$ for each $n$, so by regularity, let $x \in V_n$ and $F \subseteq U_n$ with $V_n \cap U_n = \emptyset$. Then $A \subseteq V = \bigcup_{n < \omega} V_n$ and $F \subseteq W = \bigcap_{n < \omega} U_n$ with $V \cap W = \emptyset$. But that implies $\cl{V} \cap F = \emptyset$, so $H \subseteq V \subseteq \cl{V} \subseteq U$.

If $A,B$ are closed and disjoint, just apply the theorem to $A \subseteq U = B^C$, since $A$ must be countable. Then $A \subseteq V \subseteq \cl{V} \subseteq U$ so that $B \subseteq W = \cl{V}^C$ and $V \cap W = \emptyset$.

A topology is a coarser topology of itself.

Let $\tau$ be the current topology of $X$ and $\tau' \subseteq \tau$ a metrizable topology on $X$ with metric $d$. If $x \ne y$, then $d(p, x) + d(p, y) > 0$ for all $p \in X$. So define

$$f(p) = \frac{d(p, x)}{d(p, x) + d(p, y)}$$

Then $f : (X, \tau') \to [0, 1]$ is continuous with $f(x) = 0$ and $f(y) = 1$. Since $\tau' \subseteq \tau$, $f : (X, \tau) \to [0, 1]$ is also continuous.

The space has a coarser metrizable topology, and any coarser topology of a separable space is separable as well (a dense set remains dense in coarser topologies).

We’re just dropping the “separable”.

Let $f : X \to \R$ be injective. Using the usual topology of $\R$, we can define $d(x, y) = \abs{f(x) - f(y)}$ which is clearly a metric, and under this new metrizable topology for $X$, $f$ is a homeomorphism from $X$ to $f(X) \subseteq \R$, and any subset of $\R$ has a countable dense subset. This separable metrizable topology is trivially coarser than the discrete topology.

Consider $X$ with such a separable metrizable topology. Let $A$ be a countable dense subset. To prove the result, we wish to find a function $f : X \to \mathcal{P}(A)^{\N}$ which is injective, so that $\text{card}(X) \le \text{card}(\mathcal{P}(A)^{\N}) = (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0} = \mathfrak{c}$. Such a function is $f(x) = (f_1(x), f_2(x), \dots)$ where each $f_n : X \to \mathcal{P}(A)$ is defined as $f_n(x) = B(x, 1/n) \cap A$. As $A$ is dense, $f_n(x) \ne \emptyset$ for all $n$ and all $x$. For $x \ne y$, let $n > 2d(x, y)$ so that $B(x, 1/n) \cap B(y, 1/n) = \emptyset$, and so $f_n(x) \cap f_n(y) = \emptyset \implies f(x) \ne f(y)$.

Let $(x_n)$ be a sequence with $x_n \to x$. Then $A = \{ x_n \ : \ n < \omega\} \cup \{x\}$ is compact, since for any open cover of $A$, some element $V$ nbd of $x$ must contain $x_n$ for all $n \ge n_0$ and $\{ x_n \ : \ n < n_0 \}$ is finite. But then $A$ is finite, and in a $T_1$ space, it must be discrete. So for some $n_0$, $n \ge n_0 \implies x_n \in \{x\} \implies x_n = x$.

Any converging sequence is eventually constant, and this constant has to be unique.

All countable sets are closed, since they are a countable union of singletons (closed by $T_1$). If $A$ is countable and some $p \in A$ were not to be isolated, then $p \in \cl{A \setminus \{p\}}$, which would imply $A \setminus \{p\}$ is countable yet not closed.

Contrapositively, if $p$ is not an isolated point, then $p \in \cl{X \setminus \{p\}}$ and then there exists a sequence $(x_n)$ in $X \setminus \{p\}$ with $x_n \to p$. But then this sequence cannot be eventually constant, since the constant would have to be $p$ itself.

Let $A$ countable with $\cl{A} = X$. For any $x \in X$, $A \cup \{x\}$ is countable, so it is discrete. Since $x \in \cl{A}$, we have $x \in A$ and so $X = A$ is countable.

Contrapositively, if $p$ is not an isolated point, then $p \in \cl{X \setminus \{p\}}$, so there exists a countable $D \subseteq X \setminus \{p\}$ with $p \in \cl{D}$. But then $D \cup \{p\}$ is countable yet not discrete.

A regular open neighborhood is… an open neighborhood.

If $x \in U$ and $y \in V$ with $U \cap V = \emptyset$, in particular, $y \notin \cl{U}$. So $x \in W = \text{int}(\cl{U})$ is regular and $y \notin W$. (blaze it)

If $x \in U$ is a nbd with $y \notin U$, let $O$ be a regular basis element with $x \in O \subseteq U$.

Let $x \in U = \text{int}(\cl{U})$ with $y \notin U$. If $U$ were to be dense, we’d have $\cl{U} = X$, but $\text{int}(\cl{U}) = U \ne X = \text{int}(X)$.

Let $x \in U \subseteq K$ and $y \in V \subseteq K'$ with $U,V$ open, $K,K'$ compact, and $x \ne y$. Then $L = K \cup K'$ is compact, so it is $T_2$. Thus, let $x \in W$ and $y \in W'$ be nbds such that $W \cap W' \cap L = \emptyset$. Then $(U \cap W) \cap (V \cap W') = \emptyset$ separates $x$ and $y$ by open sets in $X$.

Let $K \subseteq X$ compact and $x \notin K$. Then $K \cup \{x\}$ is compact and $T_2$. $T_2$ spaces separate points from compact sets, so there exists a nbd $V$ of $x$ with $V \cap K = \emptyset$. So $K$ is closed.

Suppose $(x_n)$ is a sequence with $x_n \to x$ and $x_n \to y$. $\omega + 1$ with the order topology is a compact space. Define $f,g : \omega + 1 \to X$ as $f(n) = g(n) = x_n$, $f(\omega) = x$, $g(\omega) = y$. Then $f$ and $g$ are continuous. By $k_2$-Hausdorff, the set $F = \{ \alpha \in \omega + 1 \ : \ f(\alpha) = g(\alpha) \}$ is closed. Clearly by definition $\omega \subseteq F$, yet $\omega$ is not closed, since $\omega \in \cl{\omega}$. So $F = \omega + 1$ and $x = y$.

Let $p$ be the dispersion point. Clearly if $x \ne y$ are points in $X \setminus \{p\}$ then there is no path from $x$ to $y$ (disconnected implies path disconnected). So the only possible non-constant paths are from $p$ to $x \ne p$. Let $f : [0, 1] \to X$ with $f(0) = p$ and $f(1) = x$. Being $T_1$, we see that $f^{-1}(p)$ is closed, so let $t = \sup f^{-1}(p)$ and then $f(t) = p$. But for any $s > t$, the path $f|_{[s, t]}$ lies within $X \setminus \{p\}$ so it must be constant, proving $f(s) = x$ for all $s > t$, and so $f$ is discontinuous at $t$: Take a nbd of $p$ not containing $x$, yet every open interval containing $t$ must contain some $s > t$.

Let $\{K_n\}_{n=1}^\infty$ be a cover of compact nbds with $K_n \subseteq \text{int}(K_{n+1})$. Let $\mathcal{U}$ be any open cover. For each $n$, there is a finite $\mathcal{F}_n \subseteq \mathcal{U}$ which covers $K_n$. Now we use the fact that each $K_n$ is closed by KC, and so $U_n = \text{int}(K_n) \setminus K_{n-1}$ are all open (to define it at $n=1$, denote $K_0 = \emptyset$). Thus, each $U_n$ is covered by the finite sets $\mathcal{F}_n$, and if we define $\mathcal{R}_n = \{ U_n \cap F \ : \ F \in \mathcal{F}_n \}$, we get that $\mathcal{R} = \bigcup_{n=1}^\infty \mathcal{R}_n$ is an open refinement, where for each $x$, take the smallest $n$ for which $x \in \text{int}(K_n)$ so that $x \in U_n$ and so $x$ intersects at least one and at most all (finitely many) elements of $\mathcal{R}_n$.

If you have at least 3 apples, then you have at least 2 apples.

If $X$ has one or two points, clearly $X \setminus \{p\}$ has either 1 point (trivially totally disconnected), or it’s empty (also totally disconnected, since there are no connected components).

If you have at least 4 apples, then you have at least 3 apples.

If you have an infinite amount of apples, then you have at least 4 apples.

Vacuously true: there are no disjoint subsets with at least two points each.

$\{x\}$ is homeomorphic to $\R^0$.

Some nbd around the isolated point has to be homeomorphic to $\R^0$. But that implies every point has a nbd homeomorphic to $\R^0$, and so every point is isolated.

Initially, for each $x \in X$, if a sequence $(y_n)$ does not converge to $x$, then there exists a nbd $V$ of $x$ such that for all $n_0 \in \omega$, there exists some $n > n_0$ such that $y_n \notin V$. This means there are infinitely many $n$ for which $y_n \notin V$, and so we can construct a subsequence $(y_{n_k})$ such that (1) it doesn’t have a converging subsequence either and (2) $y_{n_k} \notin V$ for all $k$.

Let $X = \{x_n\}$. Contrapositively, let $\mathbf{y} = (y_n)$ be a sequence with no converging subsequence. Let $V_0$ be a nbd of $x_0$ and construct a subsequence $\mathbf{y}_0$ of $\mathbf{y}$ such that every term in $\mathbf{y}_0$ lies outside $V_0$. Inductively, given $\mathbf{y}_n$, choose $V_{n+1}$ a nbd of $x_{n+1}$, and let $\mathbf{y}_{n+1}$ be a subsequence of $\mathbf{y}_n$ which lies entirely outside of $V_{n+1}$. Note that by construction, each $\mathbf{y}_n$ is a subsequence of $\mathbf{y}_k$ for any $k < n$, so all of its terms lie outside of $\bigcup_{k \le n} V_k$. This way, we’ve constructed an open cover $\{V_n\}$ of $X$ with no finite subcover, and so $X$ is not compact.

A space is a subspace of itself.

For each $A \subseteq X$, as it is separable subspace, there is a countable $D \subseteq A$ with $A \subseteq \cl{D}$. So then $\cl{A} \subseteq \cl{\cl{D}} = \cl{D}$.

Similar proof to (T427). Let $\{K_n\}$ be an increasing sequence of compact sets which cover $X$. Let $\mathcal{U}$ be an open cover. For each $n$, there exists a finite subset $\mathcal{F}_n \subseteq \mathcal{U}$ such that $\mathcal{F}_n$ covers $K_n$. Then define $\mathcal{R}_0 = \mathcal{F}_0$ and $\mathcal{R}_{n+1} = \{ U \setminus K_n \ : \ U \in \mathcal{F}_n \}$. Clearly by construction, $\mathcal{R} = \bigcup_{n < \omega} \mathcal{R}_n$ must be an open refinement of $\mathcal{U}$. And for each $x$, let $n$ be the smallest index so that $x \in K_n$. This immediately implies $x \notin U$ for any $\mathcal{R}_k$ with $k > n$, and so $x$ intersects at most the nbds in $\bigcup_{k \le n} \mathcal{R}_n$, which is finite.

If $X$ is disconnected, let $A$ be a clopen set with $p \in A$ and $q \notin A$. Then the function $f : X \to X$ defined as $f(x) = q$ at $x \in A$ and $f(x) = p$ at $x \notin A$, then $f$ is continuous yet has no fixed point.

If $g \ne 1$, then the map $x \mapsto gx$ is continuous yet has no fixed point.

Firstly, note that a compact LOTS space must have minimum and maximum elements. If it didn’t have a maximum, the open intervals $({\leftarrow}, x)$ would be an open cover with no finite subcover. Similarly for the minimum. So let $x_0$ be the minimum and $y_0$ the maximum elements.

We first see that $A = \{ x \in X \ : \ f(x) > x \}$ is open. Given $x \in A$, if there is some $x < z < f(x)$, then take $U = ({\leftarrow}, z)$ and $V = (z, {\rightarrow})$. If there is no such $z$, let $U = ({\leftarrow}, f(x))$ and $V = (x, {\rightarrow})$. In both cases, we have $x \in U$, $f(x) \in V$, and every element of $V$ is strictly greater than every element of $U$. Let $W = f^{-1}(V) \cap U$. Then $y \in W \implies f(y) > y$ and so $x \in W \subseteq A$. Through a similar proof, $B = \{ x \in X \ : \ f(x) < x \}$ is open as well. Note that $A \cap B = \emptyset$, $y_0 \notin A$, and $x_0 \notin B$. Because $X$ is connected, this means $A \cup B \ne X$, and so there must be some $p \notin A \cup B$, so $f(p) = p$.

If it has a fixed point… it has… a point.

If $p \ne q$ were to be topologically indistinguishable, let $f(x) = p$ for all $x \ne p$ and $f(p) = q$. Then $f$ is continuous yet has no fixed point.

The partition would be $\{X\}$ itself, the trivial partition.

In general, a countable topology is second countable.

If $\text{card}(X) \ge \mathfrak{c} = \text{card}([0, 1])$, then there exists an injective function $f : [0, 1] \to X$, which is clearly continuous in the trivial space, and a path from $f(0)$ to $f(1)$.

By definition.

Can’t argue with that.

Can’t argue with that.

Can’t argue with that.

By definition.

If $f : X \to Y \subseteq \R^n$ is a homeomorphism, the balls of the form $B(r, 1/n)$ for $r \in \Q$ and $n \in \N$ form a countable basis for $\R^n$, so $f^{-1}(B(r, 1/n))$ forms a basis for $X$.

If $X \simeq Y \subseteq \R^n$, note that $\R^n$ is metrizable because it’s a normed vector space, and being metrizable is hereditary, so $Y \simeq X$ is metrizable.

Denote $x \sim y$ to mean $x$ and $y$ are indistinguishable. For each $x$, let $W(x)$ be its smallest nbd. Clearly $x \sim y \implies y \in W(x)$ by definition, and for each $y \not\sim x$, there is a nbd $V$ of $x$ with $y \notin V$, so $y \notin W(x)$. Furthermore, this proves that $\sim$ is an equivalence relation, since $x \sim y \iff W(x) = W(y)$ and the sets $[x]_{\sim} = W(x)$ form a partition of $X$.

If $\{X_i\}_{i \in I}$ is the partition for the topology of $X$, then for each $x \in X$, the set $X_i$ such that $x \in X_i$ is the smallest nbd of $x$.

Any set of the partition is a clopen set. So if $X$ is the only nonempty clopen set, the partition is $\{X\}$.

As the partition forms a basis for $X$, it must refine any open cover, and each element of the partition is a clopen set.

A space is a subspace of itself (and is homeomorphic to itself).

Being a W-space is hereditary.

By definition.

By definition.

By definition.

By definition.

By one of the equivalent definitions.

It’s an order topology.

It has a bijection $f: X \to n$ if finite, or $f : X \to \omega$ if infinite. It’s a homeomorphism either way.

We can cover the space with countable sets, and take a countable subcover.