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Continuity of countable projection from non-first countable topological space
The Next CEO of Stack OverflowProof that product topology of subspace is same as induced product topologyEquivalence of continuous and sequential continuous implies first-countable?Compact topological space not having Countable Basis?Prove that $mathbb R/mathbb Z$ is sequential and not first countableCountable product of first countable Spaces is first countableIs Every Sequentially Compact, First Countable Space Also Compact?Sequential continuity implies continuity in the weak topology on a normed spaceWhat kind of a property implies (sequentially compact $iff$ compact)?Is this quotient space first countable?Locally Compact Hausdorff Topologial Group but not Sequential?Does a sequentially continuous function take its supremum on compacts?
$begingroup$
This might be trivial, but I just want to make sure I got this right:
Let $X$ be a metric space and $I$ an uncountable index set. Let us consider $X^I$ with the product topology (of course, the topology of $X$ is the one induced by the metric). Let $J subseteq I$ be an (infinite) countable subset of $I$. Let us also consider $X^J$ with the product topology and consider the map
$F: X^I to X^J$ given by $F((x_i)_i in I) := (x_j)_j in J$, i.e. $F$ is simply the projection from $I$ to $J$.
Note that $X^I$ is not a first countable topological space, so continuity of $F$ is not the same as sequential continuity of $F$. Now my question: Is the map $F$ continuous? It is trivially sequentially continuous and if $X^I$ was first countable, then the statement follows. However, I am not used to working with nets instead of sequences (which is necessary to check continuity of $F$, since its domain is not first countable) - so I wanted to make sure that my intuition that $F$ is continuous is correct. Intuitively, nothing can go wrong, because when I want to check continuity via nets, since I consider the same topology on both spaces, it seems to be straightforward.
Thankful for any answers on this!
general-topology product-space first-countable
$endgroup$
add a comment |
$begingroup$
This might be trivial, but I just want to make sure I got this right:
Let $X$ be a metric space and $I$ an uncountable index set. Let us consider $X^I$ with the product topology (of course, the topology of $X$ is the one induced by the metric). Let $J subseteq I$ be an (infinite) countable subset of $I$. Let us also consider $X^J$ with the product topology and consider the map
$F: X^I to X^J$ given by $F((x_i)_i in I) := (x_j)_j in J$, i.e. $F$ is simply the projection from $I$ to $J$.
Note that $X^I$ is not a first countable topological space, so continuity of $F$ is not the same as sequential continuity of $F$. Now my question: Is the map $F$ continuous? It is trivially sequentially continuous and if $X^I$ was first countable, then the statement follows. However, I am not used to working with nets instead of sequences (which is necessary to check continuity of $F$, since its domain is not first countable) - so I wanted to make sure that my intuition that $F$ is continuous is correct. Intuitively, nothing can go wrong, because when I want to check continuity via nets, since I consider the same topology on both spaces, it seems to be straightforward.
Thankful for any answers on this!
general-topology product-space first-countable
$endgroup$
add a comment |
$begingroup$
This might be trivial, but I just want to make sure I got this right:
Let $X$ be a metric space and $I$ an uncountable index set. Let us consider $X^I$ with the product topology (of course, the topology of $X$ is the one induced by the metric). Let $J subseteq I$ be an (infinite) countable subset of $I$. Let us also consider $X^J$ with the product topology and consider the map
$F: X^I to X^J$ given by $F((x_i)_i in I) := (x_j)_j in J$, i.e. $F$ is simply the projection from $I$ to $J$.
Note that $X^I$ is not a first countable topological space, so continuity of $F$ is not the same as sequential continuity of $F$. Now my question: Is the map $F$ continuous? It is trivially sequentially continuous and if $X^I$ was first countable, then the statement follows. However, I am not used to working with nets instead of sequences (which is necessary to check continuity of $F$, since its domain is not first countable) - so I wanted to make sure that my intuition that $F$ is continuous is correct. Intuitively, nothing can go wrong, because when I want to check continuity via nets, since I consider the same topology on both spaces, it seems to be straightforward.
Thankful for any answers on this!
general-topology product-space first-countable
$endgroup$
This might be trivial, but I just want to make sure I got this right:
Let $X$ be a metric space and $I$ an uncountable index set. Let us consider $X^I$ with the product topology (of course, the topology of $X$ is the one induced by the metric). Let $J subseteq I$ be an (infinite) countable subset of $I$. Let us also consider $X^J$ with the product topology and consider the map
$F: X^I to X^J$ given by $F((x_i)_i in I) := (x_j)_j in J$, i.e. $F$ is simply the projection from $I$ to $J$.
Note that $X^I$ is not a first countable topological space, so continuity of $F$ is not the same as sequential continuity of $F$. Now my question: Is the map $F$ continuous? It is trivially sequentially continuous and if $X^I$ was first countable, then the statement follows. However, I am not used to working with nets instead of sequences (which is necessary to check continuity of $F$, since its domain is not first countable) - so I wanted to make sure that my intuition that $F$ is continuous is correct. Intuitively, nothing can go wrong, because when I want to check continuity via nets, since I consider the same topology on both spaces, it seems to be straightforward.
Thankful for any answers on this!
general-topology product-space first-countable
general-topology product-space first-countable
asked Mar 27 at 18:43
MarcoMarco
607
607
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$begingroup$
Of course this map is continuous: if $pi_i$ is the projection onto the space $X_i$ then any map $F$ into a product is continuous iff all compositions with all $pi_i$ ($i$ in its index set), i.e. $ pi_i circ F$ are continuous. This is the continuity characterisation of all initial topologies (e.g. see my answer here). And for your $F$ we just have $pi_i circ F = pi_i$ for all $i in J$ trivially, so $F$ is continuous.
$endgroup$
$begingroup$
You are right. Thanks a lot!
$endgroup$
– Marco
2 days ago
add a comment |
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$begingroup$
Of course this map is continuous: if $pi_i$ is the projection onto the space $X_i$ then any map $F$ into a product is continuous iff all compositions with all $pi_i$ ($i$ in its index set), i.e. $ pi_i circ F$ are continuous. This is the continuity characterisation of all initial topologies (e.g. see my answer here). And for your $F$ we just have $pi_i circ F = pi_i$ for all $i in J$ trivially, so $F$ is continuous.
$endgroup$
$begingroup$
You are right. Thanks a lot!
$endgroup$
– Marco
2 days ago
add a comment |
$begingroup$
Of course this map is continuous: if $pi_i$ is the projection onto the space $X_i$ then any map $F$ into a product is continuous iff all compositions with all $pi_i$ ($i$ in its index set), i.e. $ pi_i circ F$ are continuous. This is the continuity characterisation of all initial topologies (e.g. see my answer here). And for your $F$ we just have $pi_i circ F = pi_i$ for all $i in J$ trivially, so $F$ is continuous.
$endgroup$
$begingroup$
You are right. Thanks a lot!
$endgroup$
– Marco
2 days ago
add a comment |
$begingroup$
Of course this map is continuous: if $pi_i$ is the projection onto the space $X_i$ then any map $F$ into a product is continuous iff all compositions with all $pi_i$ ($i$ in its index set), i.e. $ pi_i circ F$ are continuous. This is the continuity characterisation of all initial topologies (e.g. see my answer here). And for your $F$ we just have $pi_i circ F = pi_i$ for all $i in J$ trivially, so $F$ is continuous.
$endgroup$
Of course this map is continuous: if $pi_i$ is the projection onto the space $X_i$ then any map $F$ into a product is continuous iff all compositions with all $pi_i$ ($i$ in its index set), i.e. $ pi_i circ F$ are continuous. This is the continuity characterisation of all initial topologies (e.g. see my answer here). And for your $F$ we just have $pi_i circ F = pi_i$ for all $i in J$ trivially, so $F$ is continuous.
edited 2 days ago
answered 2 days ago
Henno BrandsmaHenno Brandsma
115k348124
115k348124
$begingroup$
You are right. Thanks a lot!
$endgroup$
– Marco
2 days ago
add a comment |
$begingroup$
You are right. Thanks a lot!
$endgroup$
– Marco
2 days ago
$begingroup$
You are right. Thanks a lot!
$endgroup$
– Marco
2 days ago
$begingroup$
You are right. Thanks a lot!
$endgroup$
– Marco
2 days ago
add a comment |
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