A. Lelek, ZBIORY (Sets), PZWS Warszawa 1966, pages
90-95.\bigskip The first hereditarily indecomposable
continuum has been constructed by B. Knaster in 1922. This
continuum $K$, discovered by Knaster, is just named the {\it
pseudo-arc} or {\it Knaster's continuum}. The pseudo-arc $K$
is obtained as the common part of a decreasing sequence of
continua in the plane, every of which is the union of
finitely many discs, forming a chain of sets. \medskip We
will consider chains
$$\Cal D = (D_1, D_2, \dots, D_k)$$
of discs $D_i$ in the plane, having the property that any
two consecutive disc $D_i$ and $D_{i+1}$ have interior
points in common, i.e., that their intersection --- nonempty
by assumption --- is not contained in the boundary of any of
them (Figure A). Discs $D_i$ are called {\it links}  of the
chain $\Cal D$. The union of all links of the chain $\Cal D$
is a continuum. Observe that if we delete from the chain
$\Cal D$ a link which is not an end link, i.e., a link $D_i$
such that $1 < i < k$, then the union of the remaining links
is a not connected set, having two components. \par Let two
chains of discs $\Cal D$ and $\Cal D'$ be given. We say that
the chain $\Cal D'$ {\it refines} the chain $\Cal D$ if each
link $D'_s$ of the chain $\Cal D'$ is contained in the
interior of at least one link $D_i$ of the chain $\Cal D$,
i.e., if for every $D'_s \in \Cal D'$ there exists a link
$D_i \in \Cal D$ such that $D'_s \subset D_i$ and that the
disc $D'_s$ has no common points with the boundary of the
disc $D_i$ (Figure B). \par We say that the chain $\Cal D'$
is {\it crooked} in the chain $\Cal D$ if it refines $\Cal
D$ and if, for every pair of links $D_i$ and $D_j$ of the
chain $\Cal D$ such that
\begin{equation}
i + 2 < j,
\end{equation}
and for every pair of links $D'_s$ and $D'_v$ of the chain
$\Cal D'$, intersecting the links $D_i$ and $D_j$
respectively, i.e.,
$$D_i \cap D'_s \ne \emptyset \ne D_j \cap D'_v,$$
there are, in the chain $\Cal D'$, links $D'_t$ and $D'_u$, lying between
the link $D'_s$ and $D'_v$ in the same order, i.e.,
$$s < t < u < v \quad \mbox { or } \quad s > t > u > v,$$
such that
\begin{equation}
D'_t \subset D_{j-1} \quad \mbox { and } \quad D'_u
\subset D_{i+1}.
\end{equation}
In other words, a part of the chain $\Cal D'$ which is
contained between the links $D'_s$ and $D'_v$ has to form a
fold in the chain $\Cal D$ between the links $D_i$ and $D_j$
(Figure C). The chain $\Cal D'$, to go from the link $D_i$
to the link $D_j$ has first to come to the link $D_{j-1}$,
next go back to the link $D_{i+1}$, and only after this it
can reach the link $D_j$. Such a fold must exist for every
two links $D_i$ and $D_j$ having no adjoining (i.e.
neighbor) link in common. \medskip Now, let there be given
in the plane two points $a$ and $b$, and an infinite
sequence
$$\Cal D_1, \Cal D_2, \dots , \Cal D_n, \dots $$
of chains of discs, satisfying, for every $n \in \{1, 2, \dots \}$, the
following conditions:
\begin{description}
\item [$~~~1^o$] the point $a$ belongs to the first, and the point $b$ to the
last link of the chain $\Cal D_n$;
\item [$~~~2^o$] the diameter of every disc in the chain $\Cal D_n$ is less
than $\frac 1n$;
\item [$~~~3^o$] the chain $\Cal D_{n+1}$ is crooked in the chain $\Cal D_n$.
\end{description}
\par Thus, for example, the chain $\Cal D_2$ is crooked in the chain
$\Cal D_1$, and the chain $\Cal D_3$ is crooked in the chain $\Cal D_2$ (on
Figure D only a part of the chain $\Cal D_3$ is presented). The foldings are
more and more condensed and they overlapped themselves. \medskip
Denote by $K_n$ the union of all links of the chain $D_n$. From condition
$3^o$ it follows in particular that the each next chain refines the previous
one, and therefore the continua $K_n$ form a decreasing sequence
$$K_1 \supset K_2 \supset \dots \supset K_n \supset \dots $$
\par The \h{Pseudo-arc} $K$ is defined as the common part of all continua
$K_n$, i.e.,
$$K = \bigcap \{K_n: n \in \Bbb N\}.$$
By virtue of condition $1^o$ points $a$ and $b$ belong to the continuum $K$,
thus it is nondegenerate. We shall prove that the continuum $K$ is
hereditarily indecomposable.\medskip
Let $K' \subset K$ be an arbitrary continuum. To prove that the continuum
$K$ is indecomposable it is enough to show that every continuum $Y
\subset K'$ distinct from $K$' is nowhere dense in $K'$. Consider an
arbitrary point $p \in Y$ and number $\eps > 0$. \par
Since $K' \ne Y$, hence there is a point $q \in K' \setminus Y$ which
therefore is at a distance at least $\eta > 0$ from each point $y \in Y$,
where $\eta$ is a fixed number, independent from $y$. In fact, if not, then
some points of the set would be arbitrarily close to the point $q$, and so
the point $q$ would belong to the closure of the set $Y$, which is
impossible, because the set $Y$ is closed, and it does not contain the point
$q$. Let us take a natural number $n$ so large that the inequalities are
satisfied
$$ \frac 2n < \eps \quad \mbox { and } \quad \frac 3n < \eta.$$ \par
Denote by $D_i$ and $D_j$ (with $i \le j$) the links of the chain $\Cal D_n$
whose union contains the points $p$ and $q$. Since $p \in Y$, hence
$\rho (p,q) \ge \eta$, (here $\rho$ means the metric in the plane) and thus
it follows that between the links $D_i$ and $D_j$ there are at least two
links of the chain $\Cal D_n$. Indeed, in the opposite case the links $D_i$
and $D_j$ would have a neighbor common link, and choosing suitable points
$p'$ and $q'$ in the intersections of this link with the links $D_i$ and
$D_j$ we would have
$$
\rho (p,q) \le \rho (p,p')  + \rho (p',q') + \rho (q',q) < \frac 1n +
\frac 1n + \frac 1n = \frac 3n < \eta
$$
by condition $2^o$. Thus we can assume that the indices $i$ and $j$ satisfy
the inequality (1). Further, we can assume also that the point $p$ belongs
to one, and the point $q$ to the other of the links $D_i$ and $D_j$. \par
Denote by $D'_s$ and $D'_v$ the links of the chain $\Cal D_{n+1}$ which
contain points $p$ and $q$, respectively. Additionally we assume that
$p \in D_i$. In the opposite case, i.e. if $p \in D_j$, the further part of
the proof runs in the same way (with changing of the roles of the links
$D_i$ and $D_j$ as well as $D_{i+1}$  and $D_{j-1}$). \par
So,
$$p \in D_i \cap D'_s \quad \mbox { and } \quad q \in Dj \cap D'_v,$$
whence we infer, by virtue of condition $3^o$ and the definition of folding
of chains, that there are two links $D'_t$ and $D'_u$ in the chain $\Cal
D_{n+1}$ which lie between the links $D'_s$ and $D'_v$ in the same order and
which satisfy the inclusions (2). The union of the links of the chain
$\Cal D_{n+1}$ distinct from the link $D'_u$ is a not connected set (see
above) containing the point $p$ in one component, and the point $q$ in the
other. Since the continuum $K'$ joins the points $p$ and $q$ and is
contained in the union $K_{n+1}$ of all links of the chain $\Cal D_{n+1}$,
hence $K'$ must pass thru the link $D'_u$. Thus there is a point
$$x \in K' \cap D'_u$$
(Figure E). We shall prove that the point $x$ does not belong to the
continuum $Y$. Indeed, if it would be $x \in Y$, then by the same reason as
previously for the continuum $K'$ the continuum $Y$ would have to pass thru
the link $D'_t$. Then there would exist a point
$$y \in Y \cap D'_t,$$
which would be in the link $D_{j-1}$ according to (2). But
the neighbor links $D_j$ and $D_{j-1}$ have their diameters
less than $\frac 1n$ by condition $2^o$, and therefore we
would have
$$\rho (y,q) < \frac 2n < \frac 3n < \eta,$$
which is impossible because $y \in Y$. \par So, the point
$x$ belongs to the set $K' \setminus Y$ and to the link
$D_{i+1}$ simultaneously. The neighbor links $D_i$ and
$D_{i+1}$ have diameters less than $\frac 1n$, whence it
follows that
$$\rho (p,x) < \frac 2n < \eps.$$ \par
Since $\eps$ was an arbitrary positive number, hence, in
this way we have proved in that there are points of the set
$K' \setminus Y$ which lie arbitrarily closely to the point
$p$. The point $p$ was an arbitrary point of the continuum
$Y$. Thus we have proved that the continuum $Y$ is nowhere
dense in $K'$, and the proof of hereditary indecomposability
of the continuum $K$ is finished.

An example of a chain crooked in a chain with 7 links is on
Figure F.
