Examples:: *[[Examples]]* Constructions:: *[[Constructions|Used in the construction of...]]* Specializations:: *[[Specializations]]* Generalizations:: *[[Generalizations]]* Justifications and Intuition:: *[[Justifications and Intuition]]* Properties:: *[[basis for the subspace topology]]*, [[product and subspace topologies commute]] Sufficiencies:: *[[Sufficiencies]]* Equivalences:: *[[Equivalences]]* Nonexamples:: [[order topology need not equal subspace topology]] ---- > [!definition] Definition. ([[subspace topology]]) > Let $X$ be a [[topological space]] with [[topological space|topology]] $\tau$. If $Y \subset X$, the collection $\tau_{Y}:=\{ Y \cap U : U \in \tau \}$ > is a [[topological space|topology]] on $Y$, called the **subspace topology**. > \ > If we want to be explicit, we sometimes say the elements of $\tau_{Y}$ are [[relatively open set|relatively open]] in $Y$. > [!equi[[subspace topology|relatively open]] the subspace topology by considering the intersections $Y \cap Z$ of $Y$ with [[closed set|closed sets]] $Z$ of $X$. Indeed, $Y \cap Z$ is [[closed set|closed in]] $Y$ because $Y-Y \cap Z=Y \cap \underbrace{ (X-Z) }_{ \text{open in }X } \in \tau_{Y}$. > [!justification] > We need to show $\tau_{Y}$ is indeed a [[topological space|topology]] on $Y$: >1. $Y=X \cap Y$ and $X \in \tau$, hence $Y \in \tau_{Y}$. And, $\emptyset = \emptyset \cap Y$, hence $\emptyset \in \tau_{Y}$. >2. Closure under arbitrary unions can be seen as follows: $\bigcup_{\xi}^{} (Y \cap U_{\xi})=(\bigcup_{\xi}^{}U_{\xi})\cap Y \subset Y.$ >3. Closure under finite intersections can be seen as follows: $(Y \cap U_{1}) \cap (Y \cap U_{2}) \cap \dots \cap (Y \cap U_{n})=Y \cap (U_{1} \cap \dots \cap U_{n}) \subset Y.$ > [!basicexample] Considering $[0,2]$ as a [[subspace topology|subspace]] of $\mathbb{R}$ with the [[standard topology on the real line|standard topology]]. The subset $(1,2] \subset [0,2]$ is open in $[0,2]$, because $(1,2]=[0,2] \cap (1, 3)$ and $(1,3)$ is open in $\mathbb{R}^{\text{standard}}$. ^e220a6 Consider the set $Y:=[-1,1]$ as a [[subspace topology|subspace]] of $\mathbb{R}$. Which of the following sets are [[open set|open in]] $Y$? Which are [[open set|open in]] $\mathbb{R}$? (Assume the [[standard topology on the real line|standard]]=[[order topology]] on $\mathbb{R}$.) 1. **$A:= \left\{ x : \frac{1}{2} < |x| < 1 \right\}$.** 1. This equals $\left( -1, \frac{1}{2} \right) \cup (\frac{1}{2}, 1)$, thus it is [[open set|open in]] $\mathbb{R}$ as a union of [[basis for a topology|basis elements]]. 2. We have $A=( \ \left( -1, \frac{1}{2} \right) \cup (\frac{1}{2}, 1) \ ) \cap Y$ and thus $A$ is [[open set|open in]] the [[subspace topology]]. 2. **$B=\left\{ x : \frac{1}{2} < |x| \leq 1 \right\}$.** 1. This equals $\left[ -1, \frac{1}{2} \right) \cup \left(\frac{1}{2}, 1 \right]$. This cannot be written as a union of [[basis for a topology|basis elements]], for at least one such element would need to contain $1$ — but no number greater than $1$ — and [[open interval]]s don't have maximal elements. So, $B$ is not [[open set|open in]] $\mathbb{R}$. 2. But, we have that $\left( \ ( -2, \frac{1}{2}) \cup (\frac{1}{2}, 3) \ \right) \cap Y=B$ and so $B$ is [[open set|open in]] $Y$. 3. **$C=\left\{ x : \frac{1}{2} \leq |x| < 1 \right\}$.** 1. This equals $(-1, -\frac{1}{2}] \cup \left[ \frac{1}{2}, 1 \right)$. This cannot be written as a union of [[basis for a topology|basis elements]], for at least one such element would need to contain $\frac{1}{2}$ — but no number less than $\frac{1}{2}$ — and [[open interval]]s don't have minimal elements. So, $C$ is not [[open set|open in]] $\mathbb{R}$. 2. By [[basis for the subspace topology]], 'standard' basis elements of $Y$ have the form $(a,b) \cap [-1,1]$, where $(a,b)$ an [[open interval]] in $\mathbb{R}$; the only [[half-open interval]]s permitted are those of the form $[-1, c)$ or $(d, 1]$ where $c,d \in Y$. For reasoning analogous to $(1)$, then, we cannot write $C$ as a union of [[basis for a topology|basis elements]] for $C$ as this would require a [[half-open interval]] of the form $\left[ \frac{1}{2}, \ell \right]$, $\ell > \frac{1}{2} \text{ in } Y$, to be a [[subspace topology|basis element for the subspace]]. (at least one of these is wrong) 4. **$D= \{ x : \frac{1}{2} \leq |x| \leq 1 \}$. ** 1. This equals $\left[ -1, \frac{1}{2} \right] \cup [\frac{1}{2}, 1]$. By same reasoning as $3.1$ it cannot be [[open set|open in]] $\mathbb{R}$. 2. By the reasoning of $3.2$ this cannot be [[open set|open in]] $Y$. 5. **$E=\{ x : 0<|x| < 1 \}$ and $\frac{1}{x} \notin \mathbb{N}$. ** 1. $E$ is the set $(0,1) \cut \left\{ \frac{1}{n} : n \in \mathbb{N}_{ \geq 2} \right\}$. Suppose this set is [[open set|open in]] $\mathbb{R}$. Then there exists an [[open interval]] $(0,a)$, $a \in (0,1)$, which possesses no elements of the form $\frac{1}{n}, n \in \mathbb{N}_{\geq 2}$— contradiction. 2. Assume there exists $U$ [[open set|open in]] $\mathbb{R}$ such that $E=U \cap [-1,1]$... then $U$ must exclude elements of the form $\frac{1}{x}$, $x \in \mathbb{N}$, but since $E \subset U$, $U$ must also contain an [[open interval]] of the same form $(0,a)$ as in $(1)$, creating the same contradiction as in $(1)$. ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` ---- ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```