Continuous functions vanishing at infinity on a non-locally-compact space












3












$begingroup$


Let $X$ be a topological space which is not locally-compact (e.g., an infinite-dimensional Hilbert space).
Let $C_{0}(X)$ denote the space of complex-valued, continuous functions vanishing at infinity on $X$, that is, an element $fin C_{0}(X)$ is a complex-valued, continuous function on $X$ such that,for every $epsilon>0$, there exists a compact $Ksubset X$ such that $|f(x)|leq epsilon$ outside $K$.
It is my understanding that $C_{0}(X)$ is a Banach space just as $C_{0}(Y)$ with $Y$ a locally-compact space.



However, in the second page of this article, it is stated that $C_{0}(X)$ only contains the zero function when $X$ is not locally-compact, but the statement is not proved.
On the other hand, at the end of the first page of this article, it is said that $C_{0}(X)$ may be very small (which, I guess, means that there are non-vanishing elements in it), but the statement is also not proved.
Furthermore, in the accepted answer of this question, it is given an example of a non-vanishing functions vanishing at infinity on a non-locally compact space, but nothing is said about its continuity.



Since I have not a strong background in functional analysis, I really do not know where to start to prove/disprove the previous statements, thus I would appreciate any hint, or any suggestion about references dealing with these matters explicitely (that is, by giving explicit proofs).






functional-analysis banach-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    Consider the function $fcolon ell^2to mathbb R$ given by $$f(x_1, x_2, x_3, ldots)=e^{-x_1^2}.$$ Isn't this function in $C_0(ell^2)$?
    $endgroup$
    – Giuseppe Negro
    Jan 10 at 11:00










  • $begingroup$
    I was thinking to a similar example, however, if we fix $epsilon$, then $|f(x)|geqepsilon$ whenever $x=(x_{1},x_{2},x_{3},...)$ is such that $x_{1}^{2}leq ln(epsilon)$ and $x_{j}$ is arbitrary for $j>1$, and I do not know if this is a compact set.
    $endgroup$
    – SepulzioNori
    Jan 10 at 11:30












  • $begingroup$
    Of course that set is not compact. You are totally right. Sorry, my example is stupid; even the function $$f(x_1, x_2)=e^{-x_1^2}$$ is not in $C_0(mathbb R^2)$. The right example would be $$f(x_1, x_2, ldots) = e^{-x_1^2 -x_2^2 -ldots}.$$ And now it is not so obvious that this is in $C_0(ell^2)$. Actually, I am quite sure it is not.
    $endgroup$
    – Giuseppe Negro
    Jan 10 at 11:49
















3












$begingroup$


Let $X$ be a topological space which is not locally-compact (e.g., an infinite-dimensional Hilbert space).
Let $C_{0}(X)$ denote the space of complex-valued, continuous functions vanishing at infinity on $X$, that is, an element $fin C_{0}(X)$ is a complex-valued, continuous function on $X$ such that,for every $epsilon>0$, there exists a compact $Ksubset X$ such that $|f(x)|leq epsilon$ outside $K$.
It is my understanding that $C_{0}(X)$ is a Banach space just as $C_{0}(Y)$ with $Y$ a locally-compact space.



However, in the second page of this article, it is stated that $C_{0}(X)$ only contains the zero function when $X$ is not locally-compact, but the statement is not proved.
On the other hand, at the end of the first page of this article, it is said that $C_{0}(X)$ may be very small (which, I guess, means that there are non-vanishing elements in it), but the statement is also not proved.
Furthermore, in the accepted answer of this question, it is given an example of a non-vanishing functions vanishing at infinity on a non-locally compact space, but nothing is said about its continuity.



Since I have not a strong background in functional analysis, I really do not know where to start to prove/disprove the previous statements, thus I would appreciate any hint, or any suggestion about references dealing with these matters explicitely (that is, by giving explicit proofs).






functional-analysis banach-spaces







share|cite|improve this question









$endgroup$












  • $begingroup$
    Consider the function $fcolon ell^2to mathbb R$ given by $$f(x_1, x_2, x_3, ldots)=e^{-x_1^2}.$$ Isn't this function in $C_0(ell^2)$?
    $endgroup$
    – Giuseppe Negro
    Jan 10 at 11:00










  • $begingroup$
    I was thinking to a similar example, however, if we fix $epsilon$, then $|f(x)|geqepsilon$ whenever $x=(x_{1},x_{2},x_{3},...)$ is such that $x_{1}^{2}leq ln(epsilon)$ and $x_{j}$ is arbitrary for $j>1$, and I do not know if this is a compact set.
    $endgroup$
    – SepulzioNori
    Jan 10 at 11:30












  • $begingroup$
    Of course that set is not compact. You are totally right. Sorry, my example is stupid; even the function $$f(x_1, x_2)=e^{-x_1^2}$$ is not in $C_0(mathbb R^2)$. The right example would be $$f(x_1, x_2, ldots) = e^{-x_1^2 -x_2^2 -ldots}.$$ And now it is not so obvious that this is in $C_0(ell^2)$. Actually, I am quite sure it is not.
    $endgroup$
    – Giuseppe Negro
    Jan 10 at 11:49














3












3








3


0



$begingroup$


Let $X$ be a topological space which is not locally-compact (e.g., an infinite-dimensional Hilbert space).
Let $C_{0}(X)$ denote the space of complex-valued, continuous functions vanishing at infinity on $X$, that is, an element $fin C_{0}(X)$ is a complex-valued, continuous function on $X$ such that,for every $epsilon>0$, there exists a compact $Ksubset X$ such that $|f(x)|leq epsilon$ outside $K$.
It is my understanding that $C_{0}(X)$ is a Banach space just as $C_{0}(Y)$ with $Y$ a locally-compact space.



However, in the second page of this article, it is stated that $C_{0}(X)$ only contains the zero function when $X$ is not locally-compact, but the statement is not proved.
On the other hand, at the end of the first page of this article, it is said that $C_{0}(X)$ may be very small (which, I guess, means that there are non-vanishing elements in it), but the statement is also not proved.
Furthermore, in the accepted answer of this question, it is given an example of a non-vanishing functions vanishing at infinity on a non-locally compact space, but nothing is said about its continuity.



Since I have not a strong background in functional analysis, I really do not know where to start to prove/disprove the previous statements, thus I would appreciate any hint, or any suggestion about references dealing with these matters explicitely (that is, by giving explicit proofs).






functional-analysis banach-spaces







share|cite|improve this question









$endgroup$




Let $X$ be a topological space which is not locally-compact (e.g., an infinite-dimensional Hilbert space).
Let $C_{0}(X)$ denote the space of complex-valued, continuous functions vanishing at infinity on $X$, that is, an element $fin C_{0}(X)$ is a complex-valued, continuous function on $X$ such that,for every $epsilon>0$, there exists a compact $Ksubset X$ such that $|f(x)|leq epsilon$ outside $K$.
It is my understanding that $C_{0}(X)$ is a Banach space just as $C_{0}(Y)$ with $Y$ a locally-compact space.



However, in the second page of this article, it is stated that $C_{0}(X)$ only contains the zero function when $X$ is not locally-compact, but the statement is not proved.
On the other hand, at the end of the first page of this article, it is said that $C_{0}(X)$ may be very small (which, I guess, means that there are non-vanishing elements in it), but the statement is also not proved.
Furthermore, in the accepted answer of this question, it is given an example of a non-vanishing functions vanishing at infinity on a non-locally compact space, but nothing is said about its continuity.



Since I have not a strong background in functional analysis, I really do not know where to start to prove/disprove the previous statements, thus I would appreciate any hint, or any suggestion about references dealing with these matters explicitely (that is, by giving explicit proofs).






functional-analysis banach-spaces




functional-analysis banach-spaces






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 10 at 10:53









SepulzioNoriSepulzioNori

373217




373217












  • $begingroup$
    Consider the function $fcolon ell^2to mathbb R$ given by $$f(x_1, x_2, x_3, ldots)=e^{-x_1^2}.$$ Isn't this function in $C_0(ell^2)$?
    $endgroup$
    – Giuseppe Negro
    Jan 10 at 11:00










  • $begingroup$
    I was thinking to a similar example, however, if we fix $epsilon$, then $|f(x)|geqepsilon$ whenever $x=(x_{1},x_{2},x_{3},...)$ is such that $x_{1}^{2}leq ln(epsilon)$ and $x_{j}$ is arbitrary for $j>1$, and I do not know if this is a compact set.
    $endgroup$
    – SepulzioNori
    Jan 10 at 11:30












  • $begingroup$
    Of course that set is not compact. You are totally right. Sorry, my example is stupid; even the function $$f(x_1, x_2)=e^{-x_1^2}$$ is not in $C_0(mathbb R^2)$. The right example would be $$f(x_1, x_2, ldots) = e^{-x_1^2 -x_2^2 -ldots}.$$ And now it is not so obvious that this is in $C_0(ell^2)$. Actually, I am quite sure it is not.
    $endgroup$
    – Giuseppe Negro
    Jan 10 at 11:49


















  • $begingroup$
    Consider the function $fcolon ell^2to mathbb R$ given by $$f(x_1, x_2, x_3, ldots)=e^{-x_1^2}.$$ Isn't this function in $C_0(ell^2)$?
    $endgroup$
    – Giuseppe Negro
    Jan 10 at 11:00










  • $begingroup$
    I was thinking to a similar example, however, if we fix $epsilon$, then $|f(x)|geqepsilon$ whenever $x=(x_{1},x_{2},x_{3},...)$ is such that $x_{1}^{2}leq ln(epsilon)$ and $x_{j}$ is arbitrary for $j>1$, and I do not know if this is a compact set.
    $endgroup$
    – SepulzioNori
    Jan 10 at 11:30












  • $begingroup$
    Of course that set is not compact. You are totally right. Sorry, my example is stupid; even the function $$f(x_1, x_2)=e^{-x_1^2}$$ is not in $C_0(mathbb R^2)$. The right example would be $$f(x_1, x_2, ldots) = e^{-x_1^2 -x_2^2 -ldots}.$$ And now it is not so obvious that this is in $C_0(ell^2)$. Actually, I am quite sure it is not.
    $endgroup$
    – Giuseppe Negro
    Jan 10 at 11:49
















$begingroup$
Consider the function $fcolon ell^2to mathbb R$ given by $$f(x_1, x_2, x_3, ldots)=e^{-x_1^2}.$$ Isn't this function in $C_0(ell^2)$?
$endgroup$
– Giuseppe Negro
Jan 10 at 11:00




$begingroup$
Consider the function $fcolon ell^2to mathbb R$ given by $$f(x_1, x_2, x_3, ldots)=e^{-x_1^2}.$$ Isn't this function in $C_0(ell^2)$?
$endgroup$
– Giuseppe Negro
Jan 10 at 11:00












$begingroup$
I was thinking to a similar example, however, if we fix $epsilon$, then $|f(x)|geqepsilon$ whenever $x=(x_{1},x_{2},x_{3},...)$ is such that $x_{1}^{2}leq ln(epsilon)$ and $x_{j}$ is arbitrary for $j>1$, and I do not know if this is a compact set.
$endgroup$
– SepulzioNori
Jan 10 at 11:30






$begingroup$
I was thinking to a similar example, however, if we fix $epsilon$, then $|f(x)|geqepsilon$ whenever $x=(x_{1},x_{2},x_{3},...)$ is such that $x_{1}^{2}leq ln(epsilon)$ and $x_{j}$ is arbitrary for $j>1$, and I do not know if this is a compact set.
$endgroup$
– SepulzioNori
Jan 10 at 11:30














$begingroup$
Of course that set is not compact. You are totally right. Sorry, my example is stupid; even the function $$f(x_1, x_2)=e^{-x_1^2}$$ is not in $C_0(mathbb R^2)$. The right example would be $$f(x_1, x_2, ldots) = e^{-x_1^2 -x_2^2 -ldots}.$$ And now it is not so obvious that this is in $C_0(ell^2)$. Actually, I am quite sure it is not.
$endgroup$
– Giuseppe Negro
Jan 10 at 11:49




$begingroup$
Of course that set is not compact. You are totally right. Sorry, my example is stupid; even the function $$f(x_1, x_2)=e^{-x_1^2}$$ is not in $C_0(mathbb R^2)$. The right example would be $$f(x_1, x_2, ldots) = e^{-x_1^2 -x_2^2 -ldots}.$$ And now it is not so obvious that this is in $C_0(ell^2)$. Actually, I am quite sure it is not.
$endgroup$
– Giuseppe Negro
Jan 10 at 11:49










1 Answer
1






active

oldest

votes


















2












$begingroup$

The space $C_0(X)$ can certainly be non-trivial even if $X$ is not locally compact. Just take a locally compact space $Y$, a space $Z$ that is not locally compact and let $X$ be the disjoint union of $Y$ and $Z$. Every $fin C_0(Y)$ induces a a function $Fin C_0(X)$ by setting $F|_Y=f$, $F|_Z=0$.



That $C_0(X)$ may be small if $X$ is not locally compact is not so much a rigorous statement, but gives a good intuition. If $X$ lacks compact sets, then the condition $fin C_0(X)$ is quite rigid. In the example above there are still enough compact subsets to produce some functions in $C_0(X)$.



In the special case of infinite-dimensional normed spaces, $C_0$ is indeed trivial. To see this, let $fin C_0(X)$ and $epsilon>0$. By definition there exists $Ksubset X$ compact such that $|f|leq epsilon$ on $Xsetminus K$. By Riesz's lemma, $K$ has empty interior. Thus $|f|leq epsilon$ everywhere. Since $epsilon>0$ was arbitrary, we conclude $f=0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    A very nice and exhaustive answer, thank you very much.
    $endgroup$
    – SepulzioNori
    Jan 10 at 15:53










  • $begingroup$
    I agree, great answer, thank you.
    $endgroup$
    – Giuseppe Negro
    Jan 10 at 17:20












Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3068492%2fcontinuous-functions-vanishing-at-infinity-on-a-non-locally-compact-space%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

The space $C_0(X)$ can certainly be non-trivial even if $X$ is not locally compact. Just take a locally compact space $Y$, a space $Z$ that is not locally compact and let $X$ be the disjoint union of $Y$ and $Z$. Every $fin C_0(Y)$ induces a a function $Fin C_0(X)$ by setting $F|_Y=f$, $F|_Z=0$.



That $C_0(X)$ may be small if $X$ is not locally compact is not so much a rigorous statement, but gives a good intuition. If $X$ lacks compact sets, then the condition $fin C_0(X)$ is quite rigid. In the example above there are still enough compact subsets to produce some functions in $C_0(X)$.



In the special case of infinite-dimensional normed spaces, $C_0$ is indeed trivial. To see this, let $fin C_0(X)$ and $epsilon>0$. By definition there exists $Ksubset X$ compact such that $|f|leq epsilon$ on $Xsetminus K$. By Riesz's lemma, $K$ has empty interior. Thus $|f|leq epsilon$ everywhere. Since $epsilon>0$ was arbitrary, we conclude $f=0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    A very nice and exhaustive answer, thank you very much.
    $endgroup$
    – SepulzioNori
    Jan 10 at 15:53










  • $begingroup$
    I agree, great answer, thank you.
    $endgroup$
    – Giuseppe Negro
    Jan 10 at 17:20
















2












$begingroup$

The space $C_0(X)$ can certainly be non-trivial even if $X$ is not locally compact. Just take a locally compact space $Y$, a space $Z$ that is not locally compact and let $X$ be the disjoint union of $Y$ and $Z$. Every $fin C_0(Y)$ induces a a function $Fin C_0(X)$ by setting $F|_Y=f$, $F|_Z=0$.



That $C_0(X)$ may be small if $X$ is not locally compact is not so much a rigorous statement, but gives a good intuition. If $X$ lacks compact sets, then the condition $fin C_0(X)$ is quite rigid. In the example above there are still enough compact subsets to produce some functions in $C_0(X)$.



In the special case of infinite-dimensional normed spaces, $C_0$ is indeed trivial. To see this, let $fin C_0(X)$ and $epsilon>0$. By definition there exists $Ksubset X$ compact such that $|f|leq epsilon$ on $Xsetminus K$. By Riesz's lemma, $K$ has empty interior. Thus $|f|leq epsilon$ everywhere. Since $epsilon>0$ was arbitrary, we conclude $f=0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    A very nice and exhaustive answer, thank you very much.
    $endgroup$
    – SepulzioNori
    Jan 10 at 15:53










  • $begingroup$
    I agree, great answer, thank you.
    $endgroup$
    – Giuseppe Negro
    Jan 10 at 17:20














2












2








2





$begingroup$

The space $C_0(X)$ can certainly be non-trivial even if $X$ is not locally compact. Just take a locally compact space $Y$, a space $Z$ that is not locally compact and let $X$ be the disjoint union of $Y$ and $Z$. Every $fin C_0(Y)$ induces a a function $Fin C_0(X)$ by setting $F|_Y=f$, $F|_Z=0$.



That $C_0(X)$ may be small if $X$ is not locally compact is not so much a rigorous statement, but gives a good intuition. If $X$ lacks compact sets, then the condition $fin C_0(X)$ is quite rigid. In the example above there are still enough compact subsets to produce some functions in $C_0(X)$.



In the special case of infinite-dimensional normed spaces, $C_0$ is indeed trivial. To see this, let $fin C_0(X)$ and $epsilon>0$. By definition there exists $Ksubset X$ compact such that $|f|leq epsilon$ on $Xsetminus K$. By Riesz's lemma, $K$ has empty interior. Thus $|f|leq epsilon$ everywhere. Since $epsilon>0$ was arbitrary, we conclude $f=0$.






share|cite|improve this answer









$endgroup$



The space $C_0(X)$ can certainly be non-trivial even if $X$ is not locally compact. Just take a locally compact space $Y$, a space $Z$ that is not locally compact and let $X$ be the disjoint union of $Y$ and $Z$. Every $fin C_0(Y)$ induces a a function $Fin C_0(X)$ by setting $F|_Y=f$, $F|_Z=0$.



That $C_0(X)$ may be small if $X$ is not locally compact is not so much a rigorous statement, but gives a good intuition. If $X$ lacks compact sets, then the condition $fin C_0(X)$ is quite rigid. In the example above there are still enough compact subsets to produce some functions in $C_0(X)$.



In the special case of infinite-dimensional normed spaces, $C_0$ is indeed trivial. To see this, let $fin C_0(X)$ and $epsilon>0$. By definition there exists $Ksubset X$ compact such that $|f|leq epsilon$ on $Xsetminus K$. By Riesz's lemma, $K$ has empty interior. Thus $|f|leq epsilon$ everywhere. Since $epsilon>0$ was arbitrary, we conclude $f=0$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 10 at 13:38









MaoWaoMaoWao

3,953618




3,953618












  • $begingroup$
    A very nice and exhaustive answer, thank you very much.
    $endgroup$
    – SepulzioNori
    Jan 10 at 15:53










  • $begingroup$
    I agree, great answer, thank you.
    $endgroup$
    – Giuseppe Negro
    Jan 10 at 17:20


















  • $begingroup$
    A very nice and exhaustive answer, thank you very much.
    $endgroup$
    – SepulzioNori
    Jan 10 at 15:53










  • $begingroup$
    I agree, great answer, thank you.
    $endgroup$
    – Giuseppe Negro
    Jan 10 at 17:20
















$begingroup$
A very nice and exhaustive answer, thank you very much.
$endgroup$
– SepulzioNori
Jan 10 at 15:53




$begingroup$
A very nice and exhaustive answer, thank you very much.
$endgroup$
– SepulzioNori
Jan 10 at 15:53












$begingroup$
I agree, great answer, thank you.
$endgroup$
– Giuseppe Negro
Jan 10 at 17:20




$begingroup$
I agree, great answer, thank you.
$endgroup$
– Giuseppe Negro
Jan 10 at 17:20


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3068492%2fcontinuous-functions-vanishing-at-infinity-on-a-non-locally-compact-space%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Bressuire

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
Read more

Cabo Verde

Image
  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º) ...
Read more

Gyllenstierna

Image
.mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output .infobox.bordered .mergedrow td,.mw-parser-output .infobox.bordered .mergedrow th{border:0;border-right:1px solid ...
Read more