Connectedness

Let $(L,\le )$ be a linearly ordered set (Partial order and partially ordered set): a poset in which any two elements are comparable. Write $<$ for the associated strict order.

Rays, intervals, and the order topology

For $a\in L$ the open rays at $a$ are

$L_{<a}:=\{t\in L:t<a\},L_{>a}:=\{t\in L:a<t\},$

and ${\mathcal{S}}_L:=\{L_{<a}:a\in L\}\cup \{L_{>a}:a\in L\}$ is the family of all of them. The order topology on $L$ is

${\mathcal{T}}_{<}:=⟨{\mathcal{S}}_L⟩,$

the topology generated by ${\mathcal{S}}_L$ (Basis and subbasis for a topology, and the topology generated by a family of sets, Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison). A linearly ordered topological space is a linearly ordered set carrying its order topology. For $a,b\in L$ write

$(a,b):=\{t\in L:a<t<b\},[a,b]:=\{t\in L:a\le t\le b\},$ $[a,b):=\{t\in L:a\le t<b\},(a,b]:=\{t\in L:a<t\le b\},$

so that $(a,b)=L_{>a}\cap L_{<b}$.

A basis, and the obligation is discharged here. By A family is a basis for a unique topology iff it covers the set and every point of an intersection of two members lies in a member inside that intersection; finite intersections of any subbasis form a basis claim 2 the intersections of finitely many members of ${\mathcal{S}}_L$ form a basis for ${\mathcal{T}}_{<}$. This library takes the empty intersection to be $L$, so $L$ itself is among them. An intersection of finitely many rays is computed by collecting the lower cuts and the upper cuts separately: since $\le$ is linear, a finite nonempty set of elements of $L$ has a greatest and a least member, so $L_{<a_1}\cap \cdots \cap L_{<a_m}=L_{<a}$ with $a$ the least of the $a_i$, and $L_{>b_1}\cap \cdots \cap L_{>b_k}=L_{>b}$ with $b$ the greatest of the $b_j$. Hence every finite intersection is $L$, an open ray, or an open interval $(b,a)=L_{>b}\cap L_{<a}$, and

${\mathcal{B}}_L:=\{L\}\cup {\mathcal{S}}_L\cup \{(a,b):a,b\in L\}$

is a basis for ${\mathcal{T}}_{<}$ (Basis and subbasis for a topology, and the topology generated by a family of sets).

What a basic neighbourhood of a point looks like. Let $x\in L$. If $x$ is neither the least nor the greatest element of $L$ (Maximal element and greatest element), then some $a<x$ and some $b>x$ exist and $x\in (a,b)$; if $x$ is least, the sets $L_{<b}$ with $b>x$ are the basic sets containing $x$ apart from $L$ itself; if $x$ is greatest, they are the sets $L_{>a}$ with $a<x$. These three cases are the only ones, and every proof below that argues at a point splits along them.

Order-convex sets. A subset $C\subseteq L$ is order-convex when

$x,z\in C\text{ and }x\le w\le z⟹w\in C\mathrm{.}$

Every ray and every one of the four interval forms above is order-convex, by transitivity of $\le$; so are $\mathrm{\varnothing }$, every singleton, and $L$.

A convention that is fixed once here. A subset $C\subseteq L$ inherits two topologies that need not agree: the subspace topology from $(L,{\mathcal{T}}_{<})$ (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace) and the order topology of the restricted order on $C$. In this library "a subspace of a linearly ordered topological space" always means the subspace topology, and the phrase "the order topology of $C$" is written in full whenever the second is meant. The two do agree when $C$ is order-convex, which is the only case used here: for order-convex $C$ the trace $L_{<a}\cap C$ is $C$ if $a$ is above every element of $C$, is $\mathrm{\varnothing }$ if $a$ is below or equal to every element of $C$, and is otherwise the ray $C_{<a}$ when $a\in C$, and $C_{<c}$ for any $c\in C$ above $a$ has the same trace description; in every case the trace of a subbasic set of $L$ is a subbasic set of $C$ or is $\mathrm{\varnothing }$ or $C$, and conversely every ray of $C$ is such a trace. The general statement, for a subset that is not order-convex, is not asserted here.

Order-density and the least upper bound property

Let $(L,\le )$ be linearly ordered.

• $L$ is order-dense (or densely ordered) when for all $x,y\in L$ with $x<y$ there is $z\in L$ with $x<z<y$. Equivalently, no element of $L$ has an immediate successor above it: there is no pair $x<y$ with $(x,y)=\mathrm{\varnothing }$.
• $L$ has the least upper bound property when every nonempty $S\subseteq L$ that has an upper bound in $L$ has a least upper bound in $L$ (Upper bound, least upper bound, and strict upper bound). A least upper bound is unique when it exists, by antisymmetry: two of them bound each other, and antisymmetry of $\le$ (Partial order and partially ordered set) forces them equal. We write $\sup S$ for it.

A linear continuum is a linearly ordered set with at least two elements that is order-dense and has the least upper bound property.

The two-element requirement is not decoration. Without it the empty ordered set and every one-point ordered set would qualify vacuously, and the theorems about linear continua elsewhere in this library would have degenerate instances whose statements say nothing. A linear continuum in the sense above is automatically infinite: two elements $x<y$ produce $z_1$ strictly between them, then $z_2$ strictly between $x$ and $z_1$, and so on, and each is new because the order is strict.

$\mathbb{R}$ is a linear continuum, and its order topology is its usual topology. The order of $\mathbb{R}$ (Order on the reals, Ordered field) is linear; $\mathbb{R}$ has at least two elements, namely $0$ and $1$; it is order-dense because $x<(x+y)\mathrm{/}2<y$ whenever $x<y$ (Ordered field); and it has the least upper bound property, which is exactly the completeness axiom (Complete ordered field (least-upper-bound property), Suprema and infima are unique, Greatest lower bound (infimum), Lower bound, bounded below, bounded set). Its order topology is the usual topology, that is the metric topology of $d_{\mathbb{R}}(s,t)=\mathrm{\mid }s-t\mathrm{\mid }$ (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ is a metric, its open balls are the intervals $(x-r,x+r)$, and it is unbounded, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not): the ball $B(x,r)$ is the interval $(x-r,x+r)$ (Open ball, closed ball and sphere in a metric space, Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length, The $\epsilon$-neighbourhood and the punctured $\epsilon$-neighbourhood of a point of $\mathbb{R}$), which is a basic set of ${\mathcal{T}}_{<}$, so every set open for the metric is open for the order; and conversely $L_{<b}$ and $L_{>a}$ are open for the metric (Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen), so every subbasic set of ${\mathcal{T}}_{<}$ is metric-open and ${\mathcal{T}}_{<}\subseteq {\mathcal{T}}_{d_{\mathbb{R}}}$ by minimality of the generated topology (Basis and subbasis for a topology, and the topology generated by a family of sets). The two topologies therefore coincide, and no second topology on $\mathbb{R}$ is being introduced.

• The dictionary with the ordinal case. The order topology on an ordinal, with the half-open intervals $(\alpha ,\beta ]$ and the initial segments $[0,\beta ]$ as a basis ↗ puts a topology on an ordinal $\gamma$ using the initial segments $[0,\beta ]$ and the half-open intervals $(\alpha ,\beta ]$ as a basis, and says in its own body that this is the general order basis rewritten so that no case analysis is needed. The two agree: $[0,\beta ]={\gamma }_{<{\beta }^{+}}$ when ${\beta }^{+}\in \gamma$ and is $\gamma$ otherwise, and $(\alpha ,\beta ]={\gamma }_{>\alpha }\cap {\gamma }_{<{\beta }^{+}}$ under the same proviso, so every basic set there is a finite intersection of rays here; conversely ${\gamma }_{<\beta }=[0,\beta ]\setminus \{\beta \}$ is the union of the sets $[0,\xi ]$ with $\xi <\beta$, and ${\gamma }_{>\alpha }$ is the union of the sets $(\alpha ,\beta ]$ with $\alpha <\beta <\gamma$, so every ray here is a union of basic sets there. The two topologies have the same open sets.

• The same dictionary for the rays presentation. The order topology on a totally ordered set, with the open rays as a subbasis, and its agreement with the usual topology of $\mathbb{R}$ presents the order topology of a totally ordered set by exactly the subbasis used above and identifies it with the usual topology of $\mathbb{R}$; the present item repeats that identification because it is the one every later proof quotes, and adds the order-convexity, order-density and least-upper-bound vocabulary that the linear-continuum theorems need.

• Why the rays and not the intervals. Taking only the open intervals $(a,b)$ as a basis fails whenever $L$ has a least or a greatest element: no interval contains the least element unless some element sits below it. The rays repair this without a case split, which is why they are the subbasis of record here.

• Order-density is not topological density. Dense, nowhere dense and codense subsets of a topological space, and the criterion by basic open sets calls a subset $A\subseteq X$ dense when $\bar{A}=X$. Order-density is a property of the ordered set itself, not of a subset, and the two words coincide only by historical accident. Where both are in play this library writes order-dense in full.

Let $(X,\mathcal{T})$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison).

• A separation of $X$ is an ordered pair $(U,V)$ of open, nonempty, disjoint subsets of $X$ with $U\cup V=X$.
• $X$ is disconnected when a separation of $X$ exists, and connected when none does.
• A subset $A\subseteq X$ is a connected subset of $X$ when the space $(A,{\mathcal{T}}_A)$ is connected, ${\mathcal{T}}_A$ being the subspace topology (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace). "Disconnected subset" is read the same way.

Since $U$ and $V$ are complementary in $X$, each of them is closed as well as open; so a separation is the same thing as a partition of $X$ into two nonempty clopen pieces (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison). The clopen subsets of $X$ are those that are both open and closed, and $\mathrm{\varnothing }$ and $X$ are always among them.

The empty space and the one-point space are connected in this library. Neither admits a separation: a separation requires two nonempty disjoint sets whose union is the whole space, and neither $\mathrm{\varnothing }$ nor a singleton can be written as such a union. So both are connected under the definition above, without any special clause. This is a live convention fork and the competing choice is recorded in Which conventions this page fixes: the empty space and the one-point space, separated sets against disjoint open sets, and what is not developed here; nothing on this page depends on which is taken except the reading of the word "connected" applied to those two spaces.

Connectedness is a property of a space, not of an ambient pair. The condition above mentions only $(X,\mathcal{T})$. When it is applied to $A\subseteq X$ it is applied to the space $(A,{\mathcal{T}}_A)$, so it does not change if $A$ is regarded as a subspace of some other space inducing the same topology on $A$; in particular a subset of $A$ is connected as a subset of $A$ exactly when it is connected as a subset of $X$, by transitivity of the subspace topology (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace). This is why "connected" may be used of a subset with no ambient space named.

Spelled out for a subset. $A\subseteq X$ is disconnected exactly when there are open $U,V\subseteq X$ with

$A\subseteq U\cup V,U\cap A\mathrm{\ne }\mathrm{\varnothing },V\cap A\mathrm{\ne }\mathrm{\varnothing },U\cap V\cap A=\mathrm{\varnothing },$

because the open sets of $(A,{\mathcal{T}}_A)$ are precisely the traces $U\cap A$. Note the last condition: it asks $U$ and $V$ to be disjoint on $A$, not in $X$. Requiring $U\cap V=\mathrm{\varnothing }$ outright is a strictly stronger demand and is a different notion.

The two-point discrete space. Write $\mathbf{2}:=\{0,1\}$ with the discrete topology (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies), in which every subset is open. A separation of $X$ is the same datum as a surjective continuous map $X\to \mathbf{2}$ (Continuity of a map of topological spaces at a point and globally): given $(U,V)$, the map sending $U$ to $0$ and $V$ to $1$ is continuous because the preimage of each of the four open subsets of $\mathbf{2}$ is one of $\mathrm{\varnothing }$, $U$, $V$, $X$; given a surjective continuous $\chi :X\to \mathbf{2}$, the pair $({\chi }^{-1}[\{0\}],{\chi }^{-1}[\{1\}])$ is a separation. This reformulation is proved as a theorem on this page and is recorded here only to name $\mathbf{2}$.

Separated sets. Two subsets $A_1,A_2\subseteq X$ are separated in $X$ when

$\overline{A_1}\cap A_2=\mathrm{\varnothing }\text{and}{A}_1\cap \overline{A_2}=\mathrm{\varnothing },$

closures taken in $X$ (Interior, closure, boundary, exterior, derived set and isolated point in a topological space, A point lies in the closure of $A$ iff every basic neighbourhood of it meets $A$; the closure is the smallest closed superset and equals $A$ together with its derived set). Separated sets are disjoint, since $A_1\subseteq \overline{A_1}$; the converse fails. This is verbatim the condition Separated sets, disconnection, and connected subset of $\mathbb{R}$ uses on the real line, transported to an arbitrary space, and the theorem relating it to the definition above is the next lemma on this page.

Totally disconnected spaces, and the empty case. The vocabulary for a space all of whose connected subsets are single points is fixed later on this page, together with the components; it is not defined here because it is stated in terms of components.

• Why "nonempty" and "disjoint" are both in the definition. Drop nonempty and every space with more than one open set is "disconnected" via $(X,\mathrm{\varnothing })$. Drop disjoint and $(X,X)$ separates every nonempty space. Drop open and every space with at least two points is separated by a point and its complement. Each of the four conditions is doing work, and the four together are the weakest demand under which the notion has the consequences proved on this page.

• Ordered pair, not unordered. A separation is written as a pair for convenience only; $(U,V)$ separates exactly when $(V,U)$ does, and no statement here distinguishes them.

• The relation to the real-line definition is proved, not assumed. Separated sets, disconnection, and connected subset of $\mathbb{R}$ defines connectedness of $E\subseteq \mathbb{R}$ by the absence of a partition into two nonempty separated sets, which is a condition on closures rather than on relatively open sets. That the two definitions agree is A subspace $A\subseteq X$ is disconnected exactly when $A=A_1\cup A_2$ with $A_1,A_2$ nonempty and separated in $X$, which is the criterion this library already uses on the real line together with The connected subspaces of $\mathbb{R}$ with its usual topology are exactly the order-convex subsets, the published characterisation transported by the identification of the two descriptions of "open in $\mathbb{R}$"; until those are proved, the two words are kept apart and no statement here quietly identifies them.

Let $(X,\mathcal{T})$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison) and let $\mathbf{2}=\{0,1\}$ carry the discrete topology (The discrete, indiscrete, cofinite, cocountable, particular-point and Sierpinski topologies). Call a map $\chi :X\to \mathbf{2}$ constant when $\chi (x)=\chi (y)$ for all $x,y\in X$.

1. The following four conditions are equivalent.

• (a) $X$ is connected: no separation of $X$ exists (Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets).
• (b) The only clopen subsets of $X$ are $\mathrm{\varnothing }$ and $X$.
• (c) Every continuous map $\chi :X\to \mathbf{2}$ is constant (Continuity of a map of topological spaces at a point and globally).
• (d) The only subsets of $X$ with empty boundary are $\mathrm{\varnothing }$ and $X$ (Interior, closure, boundary, exterior, derived set and isolated point in a topological space).

2. For $A\subseteq X$ with the subspace topology (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace), $A$ is a connected subset of $X$ if and only if the only subsets of $A$ that are clopen in $(A,{\mathcal{T}}_A)$ are $\mathrm{\varnothing }$ and $A$, if and only if every continuous map $A\to \mathbf{2}$ is constant.

Claim 2 is claim 1 applied to the space $(A,{\mathcal{T}}_A)$ and is stated separately because it is the form used in every later proof on this page: a connected set is tested by showing that a continuous two-valued function on it cannot take both values.

• Which clause is used where. Clause (c) is the workhorse: it converts a connectedness claim into a statement about functions, so it composes with continuous maps and with unions, which is what makes the theorems after it short. Clause (b) is the one to use when a candidate clopen set is already in hand. Clause (d) is stated because a boundary computation is often the quickest route in a concrete space.

• Why $\mathbf{2}$ and not an arbitrary discrete space. Any discrete space with at least two points would serve for clause (c), since a non-constant map into it composes with a retraction onto two of its points. Fixing $\mathbf{2}$ avoids having to say which two, and every use below needs no more.

• The empty space satisfies all four clauses. Its only subset is $\mathrm{\varnothing }=X$, which is clopen and has empty boundary; the unique map $\mathrm{\varnothing }\to \mathbf{2}$ is constant vacuously; and no separation exists, since a separation needs a nonempty piece. So the convention of Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets is consistent with every clause here rather than being an exception to them.

Let $(X,\mathcal{T})$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison) and let $A\subseteq X$ carry the subspace topology (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace). Write $\bar{B}$ for the closure of $B$ in $X$ (Interior, closure, boundary, exterior, derived set and isolated point in a topological space). Then $A$ is a disconnected subset of $X$ (Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets) if and only if there are sets $A_1,A_2$ with

$A=A_1\cup A_2,A_1\mathrm{\ne }\mathrm{\varnothing }\mathrm{\ne }{A}_2,\overline{A_1}\cap A_2=\mathrm{\varnothing }=A_1\cap \overline{A_2}\mathrm{.}$

Equivalently: $A$ is connected if and only if it admits no such decomposition. The two sets in such a decomposition are automatically disjoint, since $A_1\cap A_2\subseteq \overline{A_1}\cap A_2=\mathrm{\varnothing }$.

The displayed condition is the one Separated sets, disconnection, and connected subset of $\mathbb{R}$ states for subsets of $\mathbb{R}$, with the closure of $\mathbb{R}$ replaced by the closure of $X$: there a disconnection of $E$ is a pair of nonempty separated sets whose union is $E$, and $E$ is connected when none exists. That the two closures on $\mathbb{R}$ are the same operation, and hence that the two definitions agree there, is proved later on this page; nothing in the present lemma asserts it.

• Why the closures are taken in $X$ and the openness in $A$. The two halves of the criterion live in different spaces on purpose. Relative openness is not visible from $X$ alone — a set open in $A$ need not be open in $X$ — whereas the closure operator of $A$ is computed from that of $X$ by [A2]. Trading the relatively open pieces for ambiently separated ones is exactly what makes the criterion usable when only $X$ is concretely known, which is the situation in every worked example on the companion page.

• Separated is strictly stronger than disjoint, and that is what is needed. If the condition asked only for a partition into two nonempty disjoint pieces then every space with at least two points would be "disconnected". The two closure conditions are what forbid one piece from clinging to the other, and each of them is used once in the proof above.

• The hypothesis $A_i\subseteq A$ is not imposed and is automatic. Both sets appear inside a union equal to $A$, so each is contained in $A$; the statement is written without the redundant hypothesis so that it can be applied directly to a candidate pair.

Give $\mathbb{R}$ its usual topology, the metric topology of $d_{\mathbb{R}}(s,t)=\mathrm{\mid }s-t\mathrm{\mid }$ (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ is a metric, its open balls are the intervals $(x-r,x+r)$, and it is unbounded, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not), and let $E\subseteq \mathbb{R}$ carry the subspace topology (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace). Then $E$ is a connected subset of $\mathbb{R}$ (Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets) if and only if $E$ is order-convex (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length, The order topology of a linearly ordered set, with the open rays as a subbasis; order-convex sets, order-density, the least upper bound property, and linear continua), that is

$x,z\in E\text{ and }x\le w\le z⟹w\in E\mathrm{.}$

In particular each of the nine interval forms of Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length is connected, and so are $\mathrm{\varnothing }$ and every singleton.

What has to be checked, and it is not the mathematics. The characterisation itself is the published A subset of $\mathbb{R}$ is connected if and only if it is order-convex, that is, an interval, which is stated for the connectedness of Separated sets, disconnection, and connected subset of $\mathbb{R}$ — a condition phrased with the open sets of Open subset of $\mathbb{R}$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen and the closure of Interior, closure, boundary and exterior of a subset of $\mathbb{R}$. The present corollary says the same thing for the connectedness of Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets in the topological space $\mathbb{R}$. What licenses the transport is that the two descriptions of "open in $\mathbb{R}$" are the same condition word for word, which is unfolded in the proof rather than quoted.

• Nothing here re-proves the hard direction. The mathematical content — that order-convexity is exactly connectedness on the line — is A subset of $\mathbb{R}$ is connected if and only if it is order-convex, that is, an interval, whose proof uses the least upper bound property. This corollary only checks that the vocabulary of the general definition and the vocabulary of the real-line definition denote the same conditions, so that the published theorem may be quoted afterwards without a translation step each time.

• "Interval" is read as "order-convex" throughout. The published theorem records that the converse classification — that every order-convex subset of $\mathbb{R}$ is empty or one of the nine written forms — is not proved, and Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length records the same omission. The statement above is therefore written with order-convexity and not with a list of forms.

• The identification is one sentence and is deliberately not routed through a conventions remark. A dependency edge onto a remark that itself points at material developed further on would mark every consequence of this corollary as resting on later material, which would be false of everything on this page. The computation $B(x,r)=(x-r,x+r)=N_r(x)$ is short enough to carry in the open.

Let $X$ and $Y$ be topological spaces and let $f:X\to Y$ be continuous (Continuity of a map of topological spaces at a point and globally). Subsets carry the subspace topology (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace). Then:

1. Images. If $A\subseteq X$ is a connected subset of $X$ (Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets) then $f[A]$ is a connected subset of $Y$. In particular, if $X$ is connected then $f[X]$ is connected, and if $f$ is moreover surjective then $Y$ is connected.
2. Topological invariance. If $h:X\to Y$ is a homeomorphism (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological) then $X$ is connected if and only if $Y$ is. So connectedness is a topological property.

Nothing is assumed about $f$ beyond continuity: it need not be injective, open, closed or surjective. Note the direction — a continuous image of a connected space is connected, while a continuous preimage need not be, since a constant map from a disconnected space is continuous.

• Why the corestriction is the only technical point. Claim 1 is about $f[A]$ as a space, so the map that must be shown continuous is the one landing in $f[A]$, not the one landing in $Y$. The characteristic property of a subspace is exactly the tool that upgrades the second to the first, and it is the reason the proof needs no hypothesis on $f$ at all.

• The hypothesis cannot be moved to the target. If $f[X]$ is connected nothing follows about $X$: the constant map from any space whatever has a one-point image, which is connected. So claim 1 is a one-way implication and is used only in that direction below.

• What the theorem buys immediately. Any property preserved by continuous images can be checked on a convenient model. That is the whole mechanism behind the intermediate value theorem in the next item, and behind the connectedness of every path-connected space later on this page: both work by pushing a connected interval forward along a continuous map.

Let $X$ be a connected topological space (Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets) and let $f:X\to \mathbb{R}$ be continuous (Continuity of a map of topological spaces at a point and globally), $\mathbb{R}$ carrying its usual topology (The absolute value makes $\mathbb{R}$ a metric space: $d(x,y)=\mathrm{\mid }x-y\mathrm{\mid }$ is a metric, its open balls are the intervals $(x-r,x+r)$, and it is unbounded, The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Metrizable space: a topological space whose topology is induced by some metric; metrizability is topological, the metric is not). Then:

1. $f[X]$ is an order-convex subset of $\mathbb{R}$ (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length).
2. Intermediate values are attained. If $p,q\in X$ and $c\in \mathbb{R}$ satisfies $f(p)\le c\le f(q)$, then there is $x\in X$ with $f(x)=c$.

Claim 2 is the intermediate value theorem with no hypothesis on $X$ beyond connectedness: no order, no metric, no interval. The classical statement for a continuous $f:[a,b]\to \mathbb{R}$ is the special case $X=[a,b]$, that subspace being connected by The connected subspaces of $\mathbb{R}$ with its usual topology are exactly the order-convex subsets, the published characterisation transported by the identification of the two descriptions of "open in $\mathbb{R}$".

• Why $f(p)\le c\le f(q)$ and not $f(p)<c<f(q)$. Order-convexity is stated with non-strict inequalities, so the endpoints are included and the statement covers $c=f(p)$ and $c=f(q)$ without a separate clause. No assumption $f(p)\le f(q)$ is needed either: if $f(q)\le c\le f(p)$ the same argument applies with the two points exchanged.

• What is not claimed. Nothing here says that $f[X]$ is an interval in the sense of one of the nine written forms of Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length; that classification of the order-convex sets is recorded there as unproved. Nor does the corollary say anything about how many $x$ satisfy $f(x)=c$, or that such an $x$ can be found by any procedure. It is an existence statement obtained by transporting connectedness, and the witness is never exhibited.

• The hypothesis on $X$ is exactly connectedness. If $X$ is disconnected the conclusion fails at once: a separation $(U,V)$ of $X$ gives a continuous $f$ equal to $0$ on $U$ and $1$ on $V$ whose image is $\{0,1\}$, which omits every value strictly between.

Let $X$ be a topological space, let $I$ be a set and let $A_i\subseteq X$ be a connected subset of $X$ for every $i\in I$ (Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets). Subsets carry the subspace topology (Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace). Then:

1. Common point. If there is $p\in X$ with $p\in A_i$ for every $i\in I$, then ${\bigcup }_{i\in I}{A}_i$ is a connected subset of $X$.
2. Common connected core. If $A\subseteq X$ is connected and $A\cap A_i\mathrm{\ne }\mathrm{\varnothing }$ for every $i\in I$, then $A\cup {\bigcup }_{i\in I}{A}_i$ is a connected subset of $X$.

No hypothesis of any kind is imposed on the index set: $I$ may be empty, finite or infinite, and no choice principle is used, since the point $p$ in claim 1 and the set $A$ in claim 2 are given rather than selected.