About the Lindelöf degree of $(R) ^I$
up vote
2
down vote
favorite
We already know that if $(R) ^ I$ is a Lindelöf space, then $I$ is countable. Is there a generalization of this proposition? If the Lindelöf degree of $(R) ^ I$ is a cardinal $lambda$, it can be concluded that the cardinal of $I$ does not exceed $lambda$?
My first attempt was a generalization of the proof of the numerable case, but the classical argument rests in the fact that a regular Lindelöf space is automatically normal, and $(R) ^ I$ is not normal if $I$ is not countable, so a generalization in this way is not evident.
Thank you for the further answers.
general-topology
asked 2 days ago
user78594
685
add a comment |
up vote
2
down vote
favorite
We already know that if $(R) ^ I$ is a Lindelöf space, then $I$ is countable. Is there a generalization of this proposition? If the Lindelöf degree of $(R) ^ I$ is a cardinal $lambda$, it can be concluded that the cardinal of $I$ does not exceed $lambda$?
My first attempt was a generalization of the proof of the numerable case, but the classical argument rests in the fact that a regular Lindelöf space is automatically normal, and $(R) ^ I$ is not normal if $I$ is not countable, so a generalization in this way is not evident.
Thank you for the further answers.
general-topology
asked 2 days ago
user78594
685
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
We already know that if $(R) ^ I$ is a Lindelöf space, then $I$ is countable. Is there a generalization of this proposition? If the Lindelöf degree of $(R) ^ I$ is a cardinal $lambda$, it can be concluded that the cardinal of $I$ does not exceed $lambda$?
My first attempt was a generalization of the proof of the numerable case, but the classical argument rests in the fact that a regular Lindelöf space is automatically normal, and $(R) ^ I$ is not normal if $I$ is not countable, so a generalization in this way is not evident.
Thank you for the further answers.
general-topology
asked 2 days ago
user78594
685
We already know that if $(R) ^ I$ is a Lindelöf space, then $I$ is countable. Is there a generalization of this proposition? If the Lindelöf degree of $(R) ^ I$ is a cardinal $lambda$, it can be concluded that the cardinal of $I$ does not exceed $lambda$?
My first attempt was a generalization of the proof of the numerable case, but the classical argument rests in the fact that a regular Lindelöf space is automatically normal, and $(R) ^ I$ is not normal if $I$ is not countable, so a generalization in this way is not evident.
Thank you for the further answers.
general-topology
general-topology
asked 2 days ago
user78594
685
asked 2 days ago
user78594
685
asked 2 days ago
user78594
685
asked 2 days ago
user78594
685
asked 2 days ago
user78594
685
685
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
If $l(mathbb{R}^I) = kappa$, then I think $|I| ge kappa$ is pretty clear: if we'd have $|I| = lambda < kappa$, then $l(mathbb{R}^I) le w(mathbb{R}^I) = |I|cdot aleph_0 = lambda< kappa$, which is a contradiction.
It wouldn't surprise me if $l(mathbb{R}^I) = |I|$, for infinite $I$.
answered yesterday
Henno Brandsma
101k344107
Thanks, that's actually true. By means of the weight, we have that |$I$| cannot be less than $lambda$. So, the problem actually is an equality between the Lindelöf degree and |$I$|.
– user78594
yesterday
Can we use a similar argument now for the weight in this case? If $omega(mathbb(R) ^I) = lambda$, the inequality that you already give above implies the searched equality.
– user78594
yesterday
@user78594 what do you mean? It is classical that $w(mathbb{R}^I) =|I|$ for $I$ infinite, but for Lindelöf degree it's harder, I think. It could be that $l(mathbb{R}^{aleph_2}) = aleph_1$ e.g. The weight ($aleph_2$) would then be greater than the Lindelöf degree, but for non-metrisable spaces that's not unheard of. My observation only confirms $aleph_2 = |I| ge aleph_1$, which is we already knew anyway.
– Henno Brandsma
22 hours ago
Yeah, the problem is around the Lindelöf degree itself, and, making a simple-minded comparison between this and the open problem about hereditarily Lindelöf- hereditarily separable spaces, I can guess that maybe this problem has no solution yet.
– user78594
18 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
If $l(mathbb{R}^I) = kappa$, then I think $|I| ge kappa$ is pretty clear: if we'd have $|I| = lambda < kappa$, then $l(mathbb{R}^I) le w(mathbb{R}^I) = |I|cdot aleph_0 = lambda< kappa$, which is a contradiction.
It wouldn't surprise me if $l(mathbb{R}^I) = |I|$, for infinite $I$.
answered yesterday
Henno Brandsma
101k344107
Thanks, that's actually true. By means of the weight, we have that |$I$| cannot be less than $lambda$. So, the problem actually is an equality between the Lindelöf degree and |$I$|.
– user78594
yesterday
Can we use a similar argument now for the weight in this case? If $omega(mathbb(R) ^I) = lambda$, the inequality that you already give above implies the searched equality.
– user78594
yesterday
@user78594 what do you mean? It is classical that $w(mathbb{R}^I) =|I|$ for $I$ infinite, but for Lindelöf degree it's harder, I think. It could be that $l(mathbb{R}^{aleph_2}) = aleph_1$ e.g. The weight ($aleph_2$) would then be greater than the Lindelöf degree, but for non-metrisable spaces that's not unheard of. My observation only confirms $aleph_2 = |I| ge aleph_1$, which is we already knew anyway.
– Henno Brandsma
22 hours ago
Yeah, the problem is around the Lindelöf degree itself, and, making a simple-minded comparison between this and the open problem about hereditarily Lindelöf- hereditarily separable spaces, I can guess that maybe this problem has no solution yet.
– user78594
18 hours ago
add a comment |
up vote
1
down vote
If $l(mathbb{R}^I) = kappa$, then I think $|I| ge kappa$ is pretty clear: if we'd have $|I| = lambda < kappa$, then $l(mathbb{R}^I) le w(mathbb{R}^I) = |I|cdot aleph_0 = lambda< kappa$, which is a contradiction.
It wouldn't surprise me if $l(mathbb{R}^I) = |I|$, for infinite $I$.
answered yesterday
Henno Brandsma
101k344107
Thanks, that's actually true. By means of the weight, we have that |$I$| cannot be less than $lambda$. So, the problem actually is an equality between the Lindelöf degree and |$I$|.
– user78594
yesterday
Can we use a similar argument now for the weight in this case? If $omega(mathbb(R) ^I) = lambda$, the inequality that you already give above implies the searched equality.
– user78594
yesterday
@user78594 what do you mean? It is classical that $w(mathbb{R}^I) =|I|$ for $I$ infinite, but for Lindelöf degree it's harder, I think. It could be that $l(mathbb{R}^{aleph_2}) = aleph_1$ e.g. The weight ($aleph_2$) would then be greater than the Lindelöf degree, but for non-metrisable spaces that's not unheard of. My observation only confirms $aleph_2 = |I| ge aleph_1$, which is we already knew anyway.
– Henno Brandsma
22 hours ago
Yeah, the problem is around the Lindelöf degree itself, and, making a simple-minded comparison between this and the open problem about hereditarily Lindelöf- hereditarily separable spaces, I can guess that maybe this problem has no solution yet.
– user78594
18 hours ago
add a comment |
up vote
1
down vote
up vote
1
down vote
If $l(mathbb{R}^I) = kappa$, then I think $|I| ge kappa$ is pretty clear: if we'd have $|I| = lambda < kappa$, then $l(mathbb{R}^I) le w(mathbb{R}^I) = |I|cdot aleph_0 = lambda< kappa$, which is a contradiction.
It wouldn't surprise me if $l(mathbb{R}^I) = |I|$, for infinite $I$.
answered yesterday
Henno Brandsma
101k344107
If $l(mathbb{R}^I) = kappa$, then I think $|I| ge kappa$ is pretty clear: if we'd have $|I| = lambda < kappa$, then $l(mathbb{R}^I) le w(mathbb{R}^I) = |I|cdot aleph_0 = lambda< kappa$, which is a contradiction.
It wouldn't surprise me if $l(mathbb{R}^I) = |I|$, for infinite $I$.
answered yesterday
Henno Brandsma
101k344107
answered yesterday
Henno Brandsma
101k344107
answered yesterday
Henno Brandsma
101k344107
answered yesterday
Henno Brandsma
101k344107
101k344107
Thanks, that's actually true. By means of the weight, we have that |$I$| cannot be less than $lambda$. So, the problem actually is an equality between the Lindelöf degree and |$I$|.
– user78594
yesterday
Can we use a similar argument now for the weight in this case? If $omega(mathbb(R) ^I) = lambda$, the inequality that you already give above implies the searched equality.
– user78594
yesterday
@user78594 what do you mean? It is classical that $w(mathbb{R}^I) =|I|$ for $I$ infinite, but for Lindelöf degree it's harder, I think. It could be that $l(mathbb{R}^{aleph_2}) = aleph_1$ e.g. The weight ($aleph_2$) would then be greater than the Lindelöf degree, but for non-metrisable spaces that's not unheard of. My observation only confirms $aleph_2 = |I| ge aleph_1$, which is we already knew anyway.
– Henno Brandsma
22 hours ago
Yeah, the problem is around the Lindelöf degree itself, and, making a simple-minded comparison between this and the open problem about hereditarily Lindelöf- hereditarily separable spaces, I can guess that maybe this problem has no solution yet.
– user78594
18 hours ago
add a comment |
Thanks, that's actually true. By means of the weight, we have that |$I$| cannot be less than $lambda$. So, the problem actually is an equality between the Lindelöf degree and |$I$|.
– user78594
yesterday
Can we use a similar argument now for the weight in this case? If $omega(mathbb(R) ^I) = lambda$, the inequality that you already give above implies the searched equality.
– user78594
yesterday
@user78594 what do you mean? It is classical that $w(mathbb{R}^I) =|I|$ for $I$ infinite, but for Lindelöf degree it's harder, I think. It could be that $l(mathbb{R}^{aleph_2}) = aleph_1$ e.g. The weight ($aleph_2$) would then be greater than the Lindelöf degree, but for non-metrisable spaces that's not unheard of. My observation only confirms $aleph_2 = |I| ge aleph_1$, which is we already knew anyway.
– Henno Brandsma
22 hours ago
Yeah, the problem is around the Lindelöf degree itself, and, making a simple-minded comparison between this and the open problem about hereditarily Lindelöf- hereditarily separable spaces, I can guess that maybe this problem has no solution yet.
– user78594
18 hours ago
Thanks, that's actually true. By means of the weight, we have that |$I$| cannot be less than $lambda$. So, the problem actually is an equality between the Lindelöf degree and |$I$|.
– user78594
yesterday
Thanks, that's actually true. By means of the weight, we have that |$I$| cannot be less than $lambda$. So, the problem actually is an equality between the Lindelöf degree and |$I$|.
– user78594
yesterday
Can we use a similar argument now for the weight in this case? If $omega(mathbb(R) ^I) = lambda$, the inequality that you already give above implies the searched equality.
– user78594
yesterday
Can we use a similar argument now for the weight in this case? If $omega(mathbb(R) ^I) = lambda$, the inequality that you already give above implies the searched equality.
– user78594
yesterday
@user78594 what do you mean? It is classical that $w(mathbb{R}^I) =|I|$ for $I$ infinite, but for Lindelöf degree it's harder, I think. It could be that $l(mathbb{R}^{aleph_2}) = aleph_1$ e.g. The weight ($aleph_2$) would then be greater than the Lindelöf degree, but for non-metrisable spaces that's not unheard of. My observation only confirms $aleph_2 = |I| ge aleph_1$, which is we already knew anyway.
– Henno Brandsma
22 hours ago
@user78594 what do you mean? It is classical that $w(mathbb{R}^I) =|I|$ for $I$ infinite, but for Lindelöf degree it's harder, I think. It could be that $l(mathbb{R}^{aleph_2}) = aleph_1$ e.g. The weight ($aleph_2$) would then be greater than the Lindelöf degree, but for non-metrisable spaces that's not unheard of. My observation only confirms $aleph_2 = |I| ge aleph_1$, which is we already knew anyway.
– Henno Brandsma
22 hours ago
Yeah, the problem is around the Lindelöf degree itself, and, making a simple-minded comparison between this and the open problem about hereditarily Lindelöf- hereditarily separable spaces, I can guess that maybe this problem has no solution yet.
– user78594
18 hours ago
Yeah, the problem is around the Lindelöf degree itself, and, making a simple-minded comparison between this and the open problem about hereditarily Lindelöf- hereditarily separable spaces, I can guess that maybe this problem has no solution yet.
– user78594
18 hours ago
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2999309%2fabout-the-lindel%25c3%25b6f-degree-of-r-i%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
