Subspaces

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

Definition of Subspaces

Just as when dealing with algebraic structures, after equipping a set \(X\) with a topological structure it is natural to examine how this structure restricts to a subset \(A\subseteq X\). The first idea that comes to mind is to select, among the open sets of \(X\), only those contained in \(A\) and declare them to be the open sets of the topological space \(A\). However, this attempt is bound to fail: if \(A\) is not an open set in \(X\), then the whole set \(A\) itself does not belong to this collection.

Definition 1 Let a topological space \((X,\mathcal{T})\) and a subset \(A\) of \(X\) be given. Then the initial topology on \(A\) defined by the inclusion \(\iota:A\hookrightarrow X\) is called the subspace topology on \(A\).

For any open set \(U\) of \(X\), we have \(\iota^{-1}(U)=U\cap A\), and for any family of open sets \((U_i)_{i\in I}\),

\[\iota^{-1}\left(\bigcup_{i\in I} U_i\right)=\left(\bigcup_{i\in I} U_i\right)\cap A=\bigcup_{i\in I} (U_i\cap A)=\bigcup_{i\in I} \iota^{-1}(U)\]

and for any finite family of open sets \((U_i)_{i\in I}\),

\[\iota^{-1}\left(\bigcap_{i\in I} U_i\right)=\left(\bigcap_{i\in I} U_i\right)\cap A=\bigcap_{i\in I} (U_i\cap A)=\bigcap_{i\in I} \iota^{-1}(U)\]

hold, so by §Initial and Final Topology, ⁋Proposition 2 we see that the subspace topology \(\mathcal{T}_A\) is given by the formula

\[\mathcal{T}_A=\{U\cap A\mid U\in\mathcal{T}\}\]

In books such as [Mun], this is sometimes taken as the definition of a subspace. From this we see that if \(A\subseteq B\subseteq X\), then whether we regard \(A\) as a subspace of \(X\) or as a subspace of \(B\) (with the subspace topology), we obtain the same space.

One must be careful when dealing with subspaces: a set may be open in \(\mathcal{T}_A\) without being open in \(\mathcal{T}\).

This situation is easily resolved if \(A\) is an open set.

Lemma 2 For a subspace \(A\) of a topological space \(X\), a necessary and sufficient condition for every open set of \(A\) to be open in \(X\) is that \(A\) be an open set in \(X\).

Proof

Since \(A\) is an open set in \(A\), one direction is trivial.

For the converse, given any open set of \(A\), there exists an open set \(U\) of \(X\) such that it can be written as \(U\cap A\); since \(A\) is also open, \(U\cap A\) is open.

It is not difficult to see that the closed sets in the subspace \(A\) are precisely the sets of the form \(A\cap C\) for closed sets \(C\) of \(X\). A similar situation can arise for closed sets, and the remedy is likewise simple.

Lemma 2 For a subspace \(A\) of a topological space \(X\), a necessary and sufficient condition for every closed set of \(A\) to be closed in \(X\) is that \(A\) be a closed set in \(X\).

Proof

Since \(A\) is a closed set in \(A\), one direction is trivial.

For the converse, given any closed set of \(A\), there exists a closed set \(U\) of \(X\) such that it can be written as \(U\cap A\); since \(A\) is also closed, \(U\cap A\) is closed.

We have defined all topological properties in terms of neighborhoods up to now, and this too can be adjusted just as in the lemmas above.

Lemma 3 For a subspace \(A\) of a topological space \(X\) and any point \(x\in A\), a necessary and sufficient condition for every neighborhood of \(x\) in \(A\) to be a neighborhood of \(x\) in \(X\) is that \(A\) be a neighborhood of \(x\) in \(X\).

Proof

Since \(A\) is a neighborhood of \(x\) in \(A\), one direction is trivial.

For the converse, let \(U\) be an arbitrary neighborhood of \(x\) in \(A\). Then there exists an open neighborhood \(U'\) of \(x\) (in \(A\)) contained in \(U\). On the other hand, if \(A\) is a neighborhood of \(x\) in \(X\), then there exists an open neighborhood \(V\) of \(x\) (in \(X\)) contained in \(A\). Now \(U'\cap V\) is a nonempty subset, \(U'\cap V\subseteq U\), and since \(U'\cap V\) is an open neighborhood of \(x\) in \(X\), it follows that \(U\) is a neighborhood of \(x\) in \(X\).

Using these lemmas, we can restrict the concepts we have examined so far to subspaces.

Proposition 4 Let \(X\) be a topological space and let \(A\subseteq B\subseteq X\) be subsets. Then the closure of \(A\) in \(B\), denoted \(\cl_BA\), is equal to

\[\cl_BA=B\cap\cl_XA\]

Proof

For any \(x\in B\), a neighborhood of \(x\) in \(B\) is always of the form \(V\cap B\) for some suitable neighborhood \(V\) of \(x\) in \(X\). Now using \(V\cap A=(V\cap B)\cap A\) and §Interior, Closure, and Boundary, ⁋Proposition 6, we obtain the desired result.

Therefore, for \(A\subseteq B\subseteq X\), the condition that \(A\) be a dense subset of \(B\) is equivalent to

\[B=\cl_BA=B\cap\cl_XA\]

and from this in turn we see that it is equivalent to \(B\subseteq\cl_XA\).

Proposition 5 Let \(X\) be a topological space and let \((A_i)_{i\in I}\) be a collection of subsets of \(X\), and suppose that one of the following two conditions holds.

1. \(X=\bigcup_{i\in I}\interior(A_i)\), or
2. \((A_i)_{i\in I}\) is a locally finite closed covering of \(X\).

Then an arbitrary subset \(B\) of \(X\) is open (resp. closed) in \(X\) if and only if \(B\cap A_i\) is open (resp. closed) in \(A_i\) for every \(i\).

Proof

First, from the identity

\[X\setminus B\cap A_i=A_i\setminus (B\cap A_i)\]

it suffices to prove the proposition for either open sets or closed sets. Also, if \(B\) is open in \(X\) then \(B\cap A_i\) is open in \(A_i\) by definition, so the heart of the proposition is the converse.

1. Assume that \((A_i)\) satisfies the first condition, and suppose that \(B\cap A_i\) is open in \(A_i\). Regarding \(A_i\) as the whole set and \(\interior A_i\) as a subspace, we know from this that \(B\cap\interior A_i\) is open in \(\interior A_i\). Since \(\interior A_i\) is an open set, applying Lemma 2 we see that \(B\cap\interior A_i\) is open in \(X\). Therefore, from

   \[B=B\cap X=B\cap\left(\bigcup_{i\in I} \interior A_i\right)=\bigcup_{i\in I}(B\cap\interior A_i)\]

   we see that \(B\) is open.

2. Now suppose that \((A_i)\) satisfies the second condition. This time we assume that all the \(B\cap A_i\) are closed in \(A_i\). Then by Lemma 2, each \(B\cap A_i\) is a closed set in \(X\). Now \((B\cap A_i)\) is a collection of locally finite closed sets, and since \(B=\bigcup (B\cap A_i)\), by §Interior, Closure, and Boundary, ⁋Proposition 4 we conclude that \(B\) is closed.

Subspaces and Continuous Functions

Let topological spaces \(X,Y\) and a function \(f:X\rightarrow Y\) be given. Then for any set \(B\) with \(f(X)\subseteq B\subseteq Y\), the function obtained by restricting the codomain of \(f\) to \(B\) is continuous. This is trivial by Definition 1 and §Initial and Final Topology, ⁋Proposition 3.

Now suppose, in the same setting, that a subset \(A\) of \(X\) is given. Then the restriction of \(f:X\rightarrow Y\) to \(A\), denoted \(f\vert_A\), equals \(f\circ\iota\) for the inclusion \(\iota:A\hookrightarrow X\). Since this is a composition of two continuous functions, we immediately see that \(f\vert_A\) is also continuous. However, the converse does not hold in general.

Example 6 Let \(f:X\rightarrow Y\) be a function between two topological spaces that is not continuous. For any \(x\in X\), if we set \(A=\{x\}\), then \(f\vert_A\) is continuous. This is because for any open set \(U\) of \(Y\), the preimage \(f^{-1}(U)\) is always either empty or \(\{x\}\).

Instead, if the set \(A\) is a neighborhood of \(x\), then the continuity of \(f\vert_A\) at the point \(x\in X\) implies that \(f\) is continuous at \(x\). This is because by Lemma 3, a neighborhood of \(x\) in \(A\) can always be viewed as a neighborhood in \(X\). To use this argument to show that \(f\) is continuous at every point, one would have to prove that for each \(x\in X\) there is a neighborhood \(N(x)\) such that \(f\vert_{N(x)}\) is continuous; however, by the following proposition we can prove that \(f\) is continuous using even weaker information.

Proposition 7 Let \(X\) be a topological space and let \((A_i)_{i\in I}\) be a collection of subsets satisfying one of the conditions of Proposition 5. Then an arbitrary function \(f:X\rightarrow Y\) into a topological space \(Y\) is continuous if and only if all the restrictions \(f\vert_{A_i}\) are continuous.

Proof

It suffices to assume that all the \(f\vert_{A_i}\) are continuous and show that \(f\) is continuous. Let \(B\) be an arbitrary closed set of \(Y\) and set \(A=f^{-1}(B)\). Since all the \(f\vert_{A_i}\) are continuous, \((f\vert_{A_i})^{-1}(B)=A\cap A_i\) are all closed sets. Applying Proposition 5, we see that \(A\) is closed, and therefore \(f\) is continuous.