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Topology MATH 342 Spring 2026

Kyle Ormsby

Section 2.8 Week 8

Subsection Monday

Subsubsection Comparing connected and path-components

Recall that the connected components of a space \(X\) are the maximal connected subspaces of \(X\text{;}\) we write \([X]\) for the set of connected components. Similarly, the path-components of \(X\) are the maximal path-connected subspaces, and we write \(\pi_0(X)\) for the set of path-components. Our goal today is to compare these two invariants and then study a local condition under which they agree.
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Observation 2.8.1.
Let \(X\) be a topological space.
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  1. The canonical map
    \begin{equation*} \coprod_{C\in [X]} C \longrightarrow X, \qquad (C\ni x\mapsto x\in X), \end{equation*}
    is a continuous bijection.
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  2. The canonical map
    \begin{equation*} \coprod_{P\in \pi_0(X)} P \longrightarrow X, \qquad (P\ni x\mapsto x\in X), \end{equation*}
    is a continuous bijection.
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Example 2.8.2. \(\mathbb{Q}\) is neither connected nor path-connected.
The rational numbers \(\mathbb{Q}\text{,}\) with the subspace topology from \(\mathbb{R}\text{,}\) are neither connected nor path-connected.
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We first show that \(\mathbb{Q}\) is disconnected. Fix any irrational number, say \(\alpha = \sqrt{2}\text{,}\) and set
\begin{equation*} U = \mathbb{Q}\cap (-\infty,\alpha), \qquad V = \mathbb{Q}\cap (\alpha,\infty). \end{equation*}
Both \(U\) and \(V\) are nonempty (e.g., \(1\in U\) and \(2\in V\)), and they are open in \(\mathbb{Q}\) since each is the intersection of \(\mathbb{Q}\) with an open interval. Since \(\alpha\notin\mathbb{Q}\text{,}\) we have \(U\cup V = \mathbb{Q}\) and \(U\cap V = \varnothing\text{,}\) so this is a separation of \(\mathbb{Q}\text{.}\)
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Since \(\mathbb{Q}\) is disconnected, it is not connected. It therefore cannot be path-connected, by Lemma 2.8.3(3).
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Proof.
We first prove statement (1). Let \(x,y\in X\) and suppose \(x\sim_{\mathrm{path}} y\text{,}\) i.e., \(x\) and \(y\) lie in the same path-component. We must show that \(x\sim_{\mathrm{conn}} y\text{,}\) i.e., that \(x\) and \(y\) lie in the same connected component. By definition of path-component, there is a continuous map \(\gamma\colon [0,1]\to X\) with \(\gamma(0)=x\) and \(\gamma(1)=y\text{.}\) In particular, \(x,y\in\gamma([0,1])\text{.}\) Since the continuous image of the connected space \([0,1]\) is connected, the subspace \(\gamma([0,1])\subseteq X\) is connected. Therefore \(x\sim_{\mathrm{conn}} y\text{,}\) which shows the map \(\pi_0(X)\to [X]\) is well-defined and that the diagram commutes. The maps from \(X\) are surjections by construction, and it follows that the induced map \(\pi_0(X)\to [X]\) is a surjection as well.
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Statement (2) follows immediately from the surjectivity established in (1). Statement (3) follows from (2): if \(X\) is path-connected then \(\pi_0(X)\) is a singleton, so \([X]\) is also a singleton, meaning \(X\) is connected.
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Example 2.8.4. Connected but not path-connected spaces.
The long line is an example of a connected topological space that is not path-connected. The topologist’s sine curve is another such example:
\begin{equation*} \bigl\{0\bigr\}\times D^1 \cup \left\{\left(x,\,\sin\tfrac{1}{x}\right) \;\middle|\; x\gt 0\right\} \;\subseteq\; \mathbb{R}^2. \end{equation*}
In each case, \(\pi_0(X)\) and \([X]\) have different cardinalities, witnessing the strictness of the inequality in Lemma 2.8.3(2).
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Subsubsection Local connectedness

We now discuss a universal property of connectedness. Unfortunately, it is possessed only by topological spaces satisfying a local condition.
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The following definition actually encodes two definitions at once: a locally connected version and a locally path-connected version, obtained by inserting or omitting the parenthetical “path-” throughout.
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Definition 2.8.5. Local (path-)connectedness.
A topological space \(X\) is locally (path-)connected :iff, for each \(x\in X\) and each open subset \(U\subseteq X\) with \(x\in U\text{,}\) there is an open subset \(V\subseteq X\) with \(x\in V\subseteq U\) and \(V\) (path-)connected.
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Equivalently, \(X\) is locally (path-)connected if there is a basis for its topology consisting of (path-)connected open subspaces.
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Example 2.8.6.
The topological space \(\mathbb{Q}\) is neither locally connected nor locally path-connected. The topologist’s sine curve is connected but not path-connected. The topological space \(\mathbb{R}^2\smallsetminus\{0\}\) is both locally connected and locally path-connected.
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Proof.
Proof.
The path-connected case is analogous to the connected case, so we give only the latter.
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Suppose \(X\) is locally connected. Let \(U\subseteq X\) be open and let \(C\subseteq U\) be a connected component of \(U\text{.}\) We show \(C\) is open in \(X\text{.}\) Let \(x\in C\text{.}\) By local connectedness, there is a connected open subset \(V\subseteq X\) with \(x\in V\subseteq U\text{.}\) Since \(x\in V\cap C\) and both \(V\) and \(C\) are connected subsets of \(U\text{,}\) the maximality of \(C\) as a connected component gives \(V\subseteq C\text{.}\) Thus every point of \(C\) has an open neighborhood contained in \(C\text{,}\) so \(C\) is open in \(X\text{.}\)
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Conversely, suppose every connected component of every open subspace of \(X\) is open in \(X\text{.}\) Let \(x\in U\) with \(U\subseteq X\) open. Let \(C\subseteq U\) be the connected component of \(U\) containing \(x\text{.}\) Then \(C\) is open in \(X\text{,}\) and \(C\) is connected with \(x\in C\subseteq U\text{.}\) This demonstrates local connectedness.
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Proof.
Remark 2.8.11.
The topologist’s sine curve demonstrates that a subspace of a locally (path-)connected topological space need not itself be locally (path-)connected.
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Subsection Wednesday

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Subsection Friday

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