General Topology: Topological Spaces


Let $(X, \mathcal{T}_X)$ be a topological space, and let $Y$ be a subset of $X$. Then the set of all intersections of open sets in $X$ with $Y$ forms a topology $\mathcal{T}_Y$ on $Y$. Formally,

$$\mathcal{T}_Y = \left\{ Y \cap T_X : T_X \in \mathcal{T}_X \right\}.$$

This topology is called the subspace topology, and $(Y, \mathcal{T}_Y)$ is called a subspace of $(X, \mathcal{T}_X)$. For clarity, we can that $\mathcal{T}_Y$ is the topology that $Y$ inherits as a subspace of $X$. Note that an open set in $T_X$ may not intersection $T_Y$, so calling any particular set "open" in the contexts of subspaces is ambiguous. To resolve this issue, we say that a set $A$ is open in $X$ or open relative to $X$ if $A \in \mathcal{T}_X$, and likewise that $A$ is open in $Y$ or open relative to $Y$ if $A \in \mathcal{T}_Y$.


  1. Let $(Y, \mathcal{T}_Y)$ be a subspace of a topological space $(X, \mathcal{T}_X)$. Show that if $Y$ is open in $\mathcal{T}_X$, then each $T_Y \in \mathcal{T}_Y$ is also open in $\mathcal{T}_X$.

    By definition of subspace, $T_Y = Y \cap T_X$ for some $T_X \in \mathcal{T}_X$. Since the intersection of two open sets is itself open, $T_Y$ is open in $\mathcal{T}_X$.

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  2. Let $\mathcal{B}$ be a basis for a topology $\mathcal{T}$ on $X$, and let $Y \subseteq X$. Show that $\mathcal{B}_Y = \{ B \cap Y : B \in \mathcal{B} \}$ is a basis for the subspace topology on $Y$.

    Let $A \in \mathcal{T}$ and let $x \in A \cap Y$. Because $A$ is a union of basis elements, there exists some $B \in \mathcal{B}$ such that $x \in B$ and $B \subseteq A$. Then $x \in B \cap Y \subseteq A \cap Y$, which by proof 5 in the Bases section shows that $\mathcal{B}$ is a basis for the subspace topology on $Y$.

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  3. Let $X$ be a topological space, let $Y$ be a subspace of $X$, and let $A$ be a subset of $Y$. Show that the subspace topology $A$ inherits from $Y$ is the same as the one it inherits from $X$.

    Each open set in $Y$ is of the form $K \cap Y$, where $K \in \mathcal{T}_X$. Likewise, each open set in $A$ in the subspace topology inherited from $Y$ is of the form $A \cap (K \cap Y)$. Because $A$ is a subset of $Y$, $A \cap (K \cap Y) = A \cap K$, which makes it belongs to the subspace on $A$ inherited by $X$. Therefore the subspace topology on $A$ inherited from $Y$ is a subset of the subspace topology on $A$ inherited by $X$. Conversely, because each open set in the subspace on $A$ inherited from $X$ is of the form $A \cap K$, where $K$ is open in $X$. Because $A \subseteq Y$, $A \cap K = A \cap (Y \cap K)$ , therefore the subspace topology on $A$ inherited from $X$ is a subset of the subspace topology on $A$ inherited by $Y$. Therefore the two subspaces are equivalent.

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  4. Let $X$ be a set and let $\mathcal{T}_{X_1}$ and $\mathcal{T}_{X_2}$ be topologies on $X$ such that $\mathcal{T}_{X_2}$ is strictly finer than $\mathcal{T}_{X_1}$. Consider the subset $Y \subset X$ and the corresponding subset topologies $\mathcal{T}_{Y_1}$ and $\mathcal{T}_{Y_2}$. Is $\mathcal{T}_{Y_2}$ comparable to $\mathcal{T}_{Y_1}$ and, if so is it finer or strictly finer than $\mathcal{T}_{Y_1}$?

    Let $T_Y \in \mathcal{T}_{Y_1}$. Then there is some $T_X \in \mathcal{T}_{X_1}$ such that $T_Y = Y \cap T_X$. Since $T_X$ is also in $\mathcal{T}_{X_2}$, it follows that $T_Y \in \mathcal{T}_{Y_2}$. Thus $\mathcal{T}_{Y_2}$ is finer than $\mathcal{T}_{Y_1}$.

    However, consider the single point subset $Y = \{x\}$. Then the two topologies $\mathcal{T}_{Y_1}$ and $\mathcal{T}_{Y_2}$ both the trivial topology (containing only $\varnothing$ and $\{x\}$). Thus, while $\mathcal{T}_{X_2}$ is strictly finer than $\mathcal{T}_{X_1}$, it is not necessarily the case that the subset topology $\mathcal{T}_{Y_2}$ is strictly finer than $\mathcal{T}_{Y_1}$.

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