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I read from the Munkres' Topology Book that

If A is a subspace of X and B is a subspace of Y , then the product topology on $A \times B$ is the same as the topology $A \times B$ inherits as a subspace of $X \times Y$.

But I got a counter example. Let's take $X = Y = \mathbb{R}$, $A = \{1\} \cup (4,5]$, and $B = \{2\} \cup (5,7]$. $1 \times 2$ is an open set in topology $A \times B$ inherits as a subspace of $X \times Y$. Same is not open in the product topology on $A \times B$.

Did I make mistake anywhere? Can You please correct me.

anomaly
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cmi
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    What does $A = 1, (4,5]$ mean? Same question for $1 \times 2$. – Randall Nov 09 '17 at 17:19
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    Why do you think it isn't open in the product topology on $A \times B$? After all, ${ 1 }$ is open in $A$ and ${ 2 }$ is open in $B$. – Daniel Schepler Nov 09 '17 at 17:29
  • $1 \times 2 $ is not open in the product topology on $A \times B$ because $1$ is not open in the order topology $A$. and same for $2$@DanielSchepler – cmi Nov 09 '17 at 17:36
  • edited @Randall – cmi Nov 09 '17 at 17:37
  • ${1}$ IS open in $A$ because ${1} = A \cap (0,1.5)$. – Randall Nov 09 '17 at 17:38
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    no $1$ can not be open set in the order topology in $A$.@Randall – cmi Nov 09 '17 at 17:42
  • OK, I give up. You should probably specify the topology in your question. – Randall Nov 09 '17 at 17:44
  • If you do not know the answer don't downvote please..@Randall – cmi Nov 09 '17 at 17:48
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    Will happily upvote once the question is clarified: the answer depends on all the topologies involved, so no one can answer you until this is declared. (Other than the answer that "the book's claim is correct".) – Randall Nov 09 '17 at 17:50
  • But what to clarify I can not understand. – cmi Nov 09 '17 at 17:52
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    Then there's your problem. If you are giving $A$ the topology inherited from the usual topology on $\mathbb{R}$ then ${1}$ is open (in $A$) as I showed earlier. If not, then no one knows if ${1}$ is open or not because you haven't specified the topology on $X, Y, \mathbb{R}, A, B$, etc. – Randall Nov 09 '17 at 17:54
  • Topology on $X$,$Y$,$A$,$B$ are Order topology. topology on $X \times Y $ and $A \times B$ are product topology.@Randall – cmi Nov 09 '17 at 17:57
  • Isn't this the usual Euclidean topology on $\mathbb{R}$, hence the usual Euclidean topology on $\mathbb{R} \times \mathbb{R}$? – Randall Nov 09 '17 at 17:59
  • Your error MAY be in thinking that if $X$ has the order topology and $A$ is a subspace then the subspace topology is also an order topology. This may or may not be false. – Randall Nov 09 '17 at 18:00
  • Topology on $X$,$Y$ are Order topology. topology on $X \times Y $ and $A \times B$ are product topology.@ – cmi Nov 09 '17 at 18:02
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    The statement from the book does need to (implicitly) assume that $A$ and $B$ are given the subspace topologies in order for this statement to be true. But, the order topology on $A$ is not the same as the subspace topology on $A$ from the order topology on $X$. – Daniel Schepler Nov 09 '17 at 18:09
  • I did not get you. Can You please elaborate?@DanielSchepler – cmi Nov 09 '17 at 18:14

1 Answers1

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In your example: $A = \{1\} \cup (4,5]$ and $B = \{2\} \cup(5,7]$.

The set $\{1\}$ is open in $A$, as it equals $(0,2) \cap A$ and $(0,2)$ is open in $X = \mathbb{R}$. (it is not open in the order topology on the set $A$, but the question is about subspace topologies not order topologies!)

Similarly $\{2\} = (1,3) \cap B$ is open in $B$, and thus $\{(1,2)\} = \{1\} \times \{2\}$ is open in $A \times B$, and this set is also open in $A \times B$ as a subspace of $\mathbb{R} \times \mathbb{R}$: $\{(1,2)\} = ((0,2) \times (1,3)) \cap (A \times B)$ and the set $(0,2) \times (1,3)$ is open in $\mathbb{R} \times \mathbb{R}$.

I gave a very abstract answer to why it's true here, but I'll try a more direct approach here, the structure of which you see in the above example as well:

$A$ and $B$ have the subspace topologies $\mathcal{T}_A$ resp. $\mathcal{T}_B$ inherited from $X$ resp. $Y$.

Suppose $O$ is open in $A \times B$ in the product topology $\mathcal{T}_A \times \mathcal{T}_B$. So $O= \bigcup_{i \in I} (U_i \times V_i)$, where $U_i$ open in $A$, $V_i$ open in $B$, for all $i \in I$.

By the definition of the subspace topology on $A$ resp. $B$, we write $U_i = \hat{U}_i \cap A$, $V_i = \hat{V}_i \cap B$ for open $\hat{U}_i$ in $X$ and open $\hat{V}_i$ in $Y$ for all $i$.

Then $\hat{O} = \bigcup_{i \in I} (\hat{U}_i \times \hat{V}_i)$ is open in $X \times Y$ in the product topology and $$\hat{O} \cap (A \times B) = (A \times B) \cap \bigcup_{i \in I} (\hat{U}_i \times \hat{V}_i) = \bigcup_{i \in I} ((\hat{U}_i \cap A) \times (\hat{V}_i \cap B)) =\bigcup_{i \in I}(U_i \times V_i) = O$$ So $\mathcal{T}_A \times \mathcal{T}_B \subseteq (\mathcal{T}_X \times \mathcal{T}_Y)_{A \times B}$.

On the other hand, if $O$ is open in $(\mathcal{T}_X \times \mathcal{T}_Y)_{A \times B}$, we write $O = \hat{O} \cap (A \times B)$ where $\hat{O}$ open in $\mathcal{T}_X \times \mathcal{T}_Y$, so $\hat{O} = \bigcup_{i \in I} (U_i \times V_i)$, with $U_i$ open in $X$, $V_i$ open in $Y$. Again we note that

$$O = \hat{O} \cap (A \times B) = (\bigcup_{i \in I} (U_i \times V_i)) \cap A \times B = \bigcup_{i \in I} ((U_i \cap A) \times (V_i \cap B))$$ which is a union of open sets of $\mathcal{T}_A \times \mathcal{T}_B$, so in that topology. This proves the other inclusion.

Henno Brandsma
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  • yes you are correct. But where was I wrong? I must have made mistake. I want to know where I went wrong...Can you please explain. Actually I can not proceed until it is solved..@Henno Brandsma – cmi Nov 10 '17 at 01:54
  • In the usual representation of $\Bbb R$ the numbers $1$ and $2$ are members of $\Bbb R$ but not subsets of $\Bbb R$ so in the Q as written, $A$ and $B$ are not subsets of $\Bbb R.$ Do you mean to say $A={1}\cup (4,5]$ and $ B={2}\cup (5,7] $ ???? – DanielWainfleet Nov 10 '17 at 05:03
  • yes yes My bad. I meant that..@DanielWainfleet – cmi Nov 10 '17 at 06:13
  • Do u have the answer?@DanielWainfleet – cmi Nov 10 '17 at 06:15