Mixing
Question about weak topology on normed space
$begingroup$
Let $(X,|quad|)$ be a normed vector space, and let $X^prime$ be the set of all bounded linear maps on $X$. I need help to clarify these questions.
- Is it correct that the weak topology on $X$ is the topology $tau_w$ such that for each $f in X^prime$ and every open interval $I=(a,b)$, $f^{-1}(I) in tau_w$?
In other words, $tau_w $ is the topology generated by ${f^{-1}(I): I = (a,b), a,b in mathbb{R} }$?
- The definition of $x_n to tilde{x}$ in $tau_w$ means that for every $U in tau_w$ with $tilde{x} in U$ there exists $M$ such that $x_n in U$ for $n geq M$.
Why is this definition equivalent to the definition
$$
f(x_n) to f(tilde{x})?
$$
as a real number for every $fin X^{prime}$?
functional-analysis
asked Jan 12 at 22:17
mmnnmmnn
566
$endgroup$
add a comment |
$begingroup$
Let $(X,|quad|)$ be a normed vector space, and let $X^prime$ be the set of all bounded linear maps on $X$. I need help to clarify these questions.
- Is it correct that the weak topology on $X$ is the topology $tau_w$ such that for each $f in X^prime$ and every open interval $I=(a,b)$, $f^{-1}(I) in tau_w$?
In other words, $tau_w $ is the topology generated by ${f^{-1}(I): I = (a,b), a,b in mathbb{R} }$?
- The definition of $x_n to tilde{x}$ in $tau_w$ means that for every $U in tau_w$ with $tilde{x} in U$ there exists $M$ such that $x_n in U$ for $n geq M$.
Why is this definition equivalent to the definition
$$
f(x_n) to f(tilde{x})?
$$
as a real number for every $fin X^{prime}$?
functional-analysis
asked Jan 12 at 22:17
mmnnmmnn
566
$endgroup$
$begingroup$
$X'$ is usually called $X^*$ ("$X$-star").
$endgroup$
– DanielWainfleet
Jan 13 at 2:59
add a comment |
$begingroup$
Let $(X,|quad|)$ be a normed vector space, and let $X^prime$ be the set of all bounded linear maps on $X$. I need help to clarify these questions.
- Is it correct that the weak topology on $X$ is the topology $tau_w$ such that for each $f in X^prime$ and every open interval $I=(a,b)$, $f^{-1}(I) in tau_w$?
In other words, $tau_w $ is the topology generated by ${f^{-1}(I): I = (a,b), a,b in mathbb{R} }$?
- The definition of $x_n to tilde{x}$ in $tau_w$ means that for every $U in tau_w$ with $tilde{x} in U$ there exists $M$ such that $x_n in U$ for $n geq M$.
Why is this definition equivalent to the definition
$$
f(x_n) to f(tilde{x})?
$$
as a real number for every $fin X^{prime}$?
functional-analysis
asked Jan 12 at 22:17
mmnnmmnn
566
$endgroup$
Let $(X,|quad|)$ be a normed vector space, and let $X^prime$ be the set of all bounded linear maps on $X$. I need help to clarify these questions.
- Is it correct that the weak topology on $X$ is the topology $tau_w$ such that for each $f in X^prime$ and every open interval $I=(a,b)$, $f^{-1}(I) in tau_w$?
In other words, $tau_w $ is the topology generated by ${f^{-1}(I): I = (a,b), a,b in mathbb{R} }$?
- The definition of $x_n to tilde{x}$ in $tau_w$ means that for every $U in tau_w$ with $tilde{x} in U$ there exists $M$ such that $x_n in U$ for $n geq M$.
Why is this definition equivalent to the definition
$$
f(x_n) to f(tilde{x})?
$$
as a real number for every $fin X^{prime}$?
functional-analysis
functional-analysis
asked Jan 12 at 22:17
mmnnmmnn
566
asked Jan 12 at 22:17
mmnnmmnn
566
asked Jan 12 at 22:17
mmnnmmnn
566
asked Jan 12 at 22:17
mmnnmmnn
566
asked Jan 12 at 22:17
mmnnmmnn
566
566
$begingroup$
$X'$ is usually called $X^*$ ("$X$-star").
$endgroup$
– DanielWainfleet
Jan 13 at 2:59
add a comment |
$begingroup$
$X'$ is usually called $X^*$ ("$X$-star").
$endgroup$
– DanielWainfleet
Jan 13 at 2:59
$begingroup$
$X'$ is usually called $X^*$ ("$X$-star").
$endgroup$
– DanielWainfleet
Jan 13 at 2:59
$begingroup$
$X'$ is usually called $X^*$ ("$X$-star").
$endgroup$
– DanielWainfleet
Jan 13 at 2:59
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes, $f^{-1}(I)$ is always weakly open. It must be, because the weak topology is the coarsest topology that makes (norm) continuous functionals continuous. Since $I$ is open in $mathbb{R}$, the inverse image $f^{-1}(I)$ is open in the weak topology.
To ask whether such sets generate the weak topology is a slightly different question, but yes, these sets form a sub-basis for the weak topology.
As for your other question about why $x_n rightharpoonup x$ is equivalent to $f(x_n) to f(x)$ for all $f in X^*$, one direction is not hard to prove. If $x_n rightharpoonup x$, $f in X^*$, and $varepsilon > 0$, then $f^{-1}(f(x) - varepsilon, f(x) + varepsilon)$ is weakly open (as above), contains $x$, and so $x_n$ must eventually lie in this set. That is, there must be some $N$ such that
begin{align*}
n > N &implies x_n in f^{-1}(f(x) - varepsilon, f(x) + varepsilon) \
&implies f(x_n) in (f(x) - varepsilon, f(x) + varepsilon) \
&implies |f(x_n) - f(x)| < varepsilon,
end{align*}
as required.
On the other hand, let's suppose $f(x_n) to f(x)$ for all $f$. We need to show that, given any weakly open neighbourhood $U$ of $x$, the sequence $(x_n)$ eventually lies inside this neighbourhood.
Recall that $U$ is a union of sub-basic sets. In particular, $x$ must be contained in a set of the form
$$V = bigcap_{k=1}^m f_k^{-1}(I_k)$$
where $I_k$ is an open interval, and $V subseteq U$. Then, since $f_k(x_n) to x$ as $n to infty$, for all $k$, there must exist some $N_k$ such that
$$n > N_k implies f_k(x_n) in I_k implies x_n in f_k^{-1}(I_k).$$
Taking $N = max{N_1, ldots, N_k}$, we get
$$n > N implies x_n in bigcap_{k=1}^m f_k^{-1}(I_k) subseteq U.$$
answered Jan 12 at 23:13
Theo BenditTheo Bendit
18.1k12152
$endgroup$
add a comment |
$begingroup$
In 1) 'is the topology ...' is is wrong. There are many topologies with this property and weak topology is the smallest topology for which $f^{-1}(I) in tau_w$ for every open interval $I$. It is the topology for which sets of the form $f^{-1}(I)$ form a subbase. (Every open set in $tau_w$ is a union of finite intersections of sets of this type).
Suppose $x_n to x$ in $tau_w$ and $fin X'$. Let $I$ be an open interval containing $f(x)$. Then $f^{-1}(I)$ is an open set in $tau_w$ containing $x$ so it contains $x_n$ for $n$ sufficiently large. Hence $f(x_n) in I$ for such $n$ proving that $f(x_n) to f(x)$. Conversely, suppose $f(x_n) to f(x)$ for every $f in X'$. Let $V$ be a neighborhood of $x$ in $tau_w$. Then there exist $f_1,f_2,...f_k in X'$ and an open intervals $I_1,I_2,...,I_k$ such that $x in cap_j f_j^{-1}(I_j) subset V$. Hence $f_j(x) in I_j$ for each $j leq k$ which implies $f_j(x_n) in I_j$ for $n$ sufficiently large. It follows that $x_n in f_j^{-1}(I_j) subset V$ for $n$ sufficiently large. Hence $x_n to x$ in $tau_w$.
answered Jan 12 at 23:23
Kavi Rama MurthyKavi Rama Murthy
59.5k42161
$endgroup$
$begingroup$
Why is the set of the form $f^{-1}(I)$ a base? Can we only say that it is just a subbasis as the other person below said?
$endgroup$
– mmnn
Jan 14 at 15:42
$begingroup$
@mmnn My answer was faulty. I have corrected the answer.
$endgroup$
– Kavi Rama Murthy
Jan 14 at 23:26
add a comment |
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
});
}
});
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%2f3071453%2fquestion-about-weak-topology-on-normed-space%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, $f^{-1}(I)$ is always weakly open. It must be, because the weak topology is the coarsest topology that makes (norm) continuous functionals continuous. Since $I$ is open in $mathbb{R}$, the inverse image $f^{-1}(I)$ is open in the weak topology.
To ask whether such sets generate the weak topology is a slightly different question, but yes, these sets form a sub-basis for the weak topology.
As for your other question about why $x_n rightharpoonup x$ is equivalent to $f(x_n) to f(x)$ for all $f in X^*$, one direction is not hard to prove. If $x_n rightharpoonup x$, $f in X^*$, and $varepsilon > 0$, then $f^{-1}(f(x) - varepsilon, f(x) + varepsilon)$ is weakly open (as above), contains $x$, and so $x_n$ must eventually lie in this set. That is, there must be some $N$ such that
begin{align*}
n > N &implies x_n in f^{-1}(f(x) - varepsilon, f(x) + varepsilon) \
&implies f(x_n) in (f(x) - varepsilon, f(x) + varepsilon) \
&implies |f(x_n) - f(x)| < varepsilon,
end{align*}
as required.
On the other hand, let's suppose $f(x_n) to f(x)$ for all $f$. We need to show that, given any weakly open neighbourhood $U$ of $x$, the sequence $(x_n)$ eventually lies inside this neighbourhood.
Recall that $U$ is a union of sub-basic sets. In particular, $x$ must be contained in a set of the form
$$V = bigcap_{k=1}^m f_k^{-1}(I_k)$$
where $I_k$ is an open interval, and $V subseteq U$. Then, since $f_k(x_n) to x$ as $n to infty$, for all $k$, there must exist some $N_k$ such that
$$n > N_k implies f_k(x_n) in I_k implies x_n in f_k^{-1}(I_k).$$
Taking $N = max{N_1, ldots, N_k}$, we get
$$n > N implies x_n in bigcap_{k=1}^m f_k^{-1}(I_k) subseteq U.$$
answered Jan 12 at 23:13
Theo BenditTheo Bendit
18.1k12152
$endgroup$
add a comment |
$begingroup$
Yes, $f^{-1}(I)$ is always weakly open. It must be, because the weak topology is the coarsest topology that makes (norm) continuous functionals continuous. Since $I$ is open in $mathbb{R}$, the inverse image $f^{-1}(I)$ is open in the weak topology.
To ask whether such sets generate the weak topology is a slightly different question, but yes, these sets form a sub-basis for the weak topology.
As for your other question about why $x_n rightharpoonup x$ is equivalent to $f(x_n) to f(x)$ for all $f in X^*$, one direction is not hard to prove. If $x_n rightharpoonup x$, $f in X^*$, and $varepsilon > 0$, then $f^{-1}(f(x) - varepsilon, f(x) + varepsilon)$ is weakly open (as above), contains $x$, and so $x_n$ must eventually lie in this set. That is, there must be some $N$ such that
begin{align*}
n > N &implies x_n in f^{-1}(f(x) - varepsilon, f(x) + varepsilon) \
&implies f(x_n) in (f(x) - varepsilon, f(x) + varepsilon) \
&implies |f(x_n) - f(x)| < varepsilon,
end{align*}
as required.
On the other hand, let's suppose $f(x_n) to f(x)$ for all $f$. We need to show that, given any weakly open neighbourhood $U$ of $x$, the sequence $(x_n)$ eventually lies inside this neighbourhood.
Recall that $U$ is a union of sub-basic sets. In particular, $x$ must be contained in a set of the form
$$V = bigcap_{k=1}^m f_k^{-1}(I_k)$$
where $I_k$ is an open interval, and $V subseteq U$. Then, since $f_k(x_n) to x$ as $n to infty$, for all $k$, there must exist some $N_k$ such that
$$n > N_k implies f_k(x_n) in I_k implies x_n in f_k^{-1}(I_k).$$
Taking $N = max{N_1, ldots, N_k}$, we get
$$n > N implies x_n in bigcap_{k=1}^m f_k^{-1}(I_k) subseteq U.$$
answered Jan 12 at 23:13
Theo BenditTheo Bendit
18.1k12152
$endgroup$
add a comment |
$begingroup$
Yes, $f^{-1}(I)$ is always weakly open. It must be, because the weak topology is the coarsest topology that makes (norm) continuous functionals continuous. Since $I$ is open in $mathbb{R}$, the inverse image $f^{-1}(I)$ is open in the weak topology.
To ask whether such sets generate the weak topology is a slightly different question, but yes, these sets form a sub-basis for the weak topology.
As for your other question about why $x_n rightharpoonup x$ is equivalent to $f(x_n) to f(x)$ for all $f in X^*$, one direction is not hard to prove. If $x_n rightharpoonup x$, $f in X^*$, and $varepsilon > 0$, then $f^{-1}(f(x) - varepsilon, f(x) + varepsilon)$ is weakly open (as above), contains $x$, and so $x_n$ must eventually lie in this set. That is, there must be some $N$ such that
begin{align*}
n > N &implies x_n in f^{-1}(f(x) - varepsilon, f(x) + varepsilon) \
&implies f(x_n) in (f(x) - varepsilon, f(x) + varepsilon) \
&implies |f(x_n) - f(x)| < varepsilon,
end{align*}
as required.
On the other hand, let's suppose $f(x_n) to f(x)$ for all $f$. We need to show that, given any weakly open neighbourhood $U$ of $x$, the sequence $(x_n)$ eventually lies inside this neighbourhood.
Recall that $U$ is a union of sub-basic sets. In particular, $x$ must be contained in a set of the form
$$V = bigcap_{k=1}^m f_k^{-1}(I_k)$$
where $I_k$ is an open interval, and $V subseteq U$. Then, since $f_k(x_n) to x$ as $n to infty$, for all $k$, there must exist some $N_k$ such that
$$n > N_k implies f_k(x_n) in I_k implies x_n in f_k^{-1}(I_k).$$
Taking $N = max{N_1, ldots, N_k}$, we get
$$n > N implies x_n in bigcap_{k=1}^m f_k^{-1}(I_k) subseteq U.$$
answered Jan 12 at 23:13
Theo BenditTheo Bendit
18.1k12152
$endgroup$
Yes, $f^{-1}(I)$ is always weakly open. It must be, because the weak topology is the coarsest topology that makes (norm) continuous functionals continuous. Since $I$ is open in $mathbb{R}$, the inverse image $f^{-1}(I)$ is open in the weak topology.
To ask whether such sets generate the weak topology is a slightly different question, but yes, these sets form a sub-basis for the weak topology.
As for your other question about why $x_n rightharpoonup x$ is equivalent to $f(x_n) to f(x)$ for all $f in X^*$, one direction is not hard to prove. If $x_n rightharpoonup x$, $f in X^*$, and $varepsilon > 0$, then $f^{-1}(f(x) - varepsilon, f(x) + varepsilon)$ is weakly open (as above), contains $x$, and so $x_n$ must eventually lie in this set. That is, there must be some $N$ such that
begin{align*}
n > N &implies x_n in f^{-1}(f(x) - varepsilon, f(x) + varepsilon) \
&implies f(x_n) in (f(x) - varepsilon, f(x) + varepsilon) \
&implies |f(x_n) - f(x)| < varepsilon,
end{align*}
as required.
On the other hand, let's suppose $f(x_n) to f(x)$ for all $f$. We need to show that, given any weakly open neighbourhood $U$ of $x$, the sequence $(x_n)$ eventually lies inside this neighbourhood.
Recall that $U$ is a union of sub-basic sets. In particular, $x$ must be contained in a set of the form
$$V = bigcap_{k=1}^m f_k^{-1}(I_k)$$
where $I_k$ is an open interval, and $V subseteq U$. Then, since $f_k(x_n) to x$ as $n to infty$, for all $k$, there must exist some $N_k$ such that
$$n > N_k implies f_k(x_n) in I_k implies x_n in f_k^{-1}(I_k).$$
Taking $N = max{N_1, ldots, N_k}$, we get
$$n > N implies x_n in bigcap_{k=1}^m f_k^{-1}(I_k) subseteq U.$$
answered Jan 12 at 23:13
Theo BenditTheo Bendit
18.1k12152
answered Jan 12 at 23:13
Theo BenditTheo Bendit
18.1k12152
answered Jan 12 at 23:13
Theo BenditTheo Bendit
18.1k12152
answered Jan 12 at 23:13
Theo BenditTheo Bendit
18.1k12152
18.1k12152
add a comment |
add a comment |
$begingroup$
In 1) 'is the topology ...' is is wrong. There are many topologies with this property and weak topology is the smallest topology for which $f^{-1}(I) in tau_w$ for every open interval $I$. It is the topology for which sets of the form $f^{-1}(I)$ form a subbase. (Every open set in $tau_w$ is a union of finite intersections of sets of this type).
Suppose $x_n to x$ in $tau_w$ and $fin X'$. Let $I$ be an open interval containing $f(x)$. Then $f^{-1}(I)$ is an open set in $tau_w$ containing $x$ so it contains $x_n$ for $n$ sufficiently large. Hence $f(x_n) in I$ for such $n$ proving that $f(x_n) to f(x)$. Conversely, suppose $f(x_n) to f(x)$ for every $f in X'$. Let $V$ be a neighborhood of $x$ in $tau_w$. Then there exist $f_1,f_2,...f_k in X'$ and an open intervals $I_1,I_2,...,I_k$ such that $x in cap_j f_j^{-1}(I_j) subset V$. Hence $f_j(x) in I_j$ for each $j leq k$ which implies $f_j(x_n) in I_j$ for $n$ sufficiently large. It follows that $x_n in f_j^{-1}(I_j) subset V$ for $n$ sufficiently large. Hence $x_n to x$ in $tau_w$.
answered Jan 12 at 23:23
Kavi Rama MurthyKavi Rama Murthy
59.5k42161
$endgroup$
$begingroup$
Why is the set of the form $f^{-1}(I)$ a base? Can we only say that it is just a subbasis as the other person below said?
$endgroup$
– mmnn
Jan 14 at 15:42
$begingroup$
@mmnn My answer was faulty. I have corrected the answer.
$endgroup$
– Kavi Rama Murthy
Jan 14 at 23:26
add a comment |
$begingroup$
In 1) 'is the topology ...' is is wrong. There are many topologies with this property and weak topology is the smallest topology for which $f^{-1}(I) in tau_w$ for every open interval $I$. It is the topology for which sets of the form $f^{-1}(I)$ form a subbase. (Every open set in $tau_w$ is a union of finite intersections of sets of this type).
Suppose $x_n to x$ in $tau_w$ and $fin X'$. Let $I$ be an open interval containing $f(x)$. Then $f^{-1}(I)$ is an open set in $tau_w$ containing $x$ so it contains $x_n$ for $n$ sufficiently large. Hence $f(x_n) in I$ for such $n$ proving that $f(x_n) to f(x)$. Conversely, suppose $f(x_n) to f(x)$ for every $f in X'$. Let $V$ be a neighborhood of $x$ in $tau_w$. Then there exist $f_1,f_2,...f_k in X'$ and an open intervals $I_1,I_2,...,I_k$ such that $x in cap_j f_j^{-1}(I_j) subset V$. Hence $f_j(x) in I_j$ for each $j leq k$ which implies $f_j(x_n) in I_j$ for $n$ sufficiently large. It follows that $x_n in f_j^{-1}(I_j) subset V$ for $n$ sufficiently large. Hence $x_n to x$ in $tau_w$.
answered Jan 12 at 23:23
Kavi Rama MurthyKavi Rama Murthy
59.5k42161
$endgroup$
$begingroup$
Why is the set of the form $f^{-1}(I)$ a base? Can we only say that it is just a subbasis as the other person below said?
$endgroup$
– mmnn
Jan 14 at 15:42
$begingroup$
@mmnn My answer was faulty. I have corrected the answer.
$endgroup$
– Kavi Rama Murthy
Jan 14 at 23:26
add a comment |
$begingroup$
In 1) 'is the topology ...' is is wrong. There are many topologies with this property and weak topology is the smallest topology for which $f^{-1}(I) in tau_w$ for every open interval $I$. It is the topology for which sets of the form $f^{-1}(I)$ form a subbase. (Every open set in $tau_w$ is a union of finite intersections of sets of this type).
Suppose $x_n to x$ in $tau_w$ and $fin X'$. Let $I$ be an open interval containing $f(x)$. Then $f^{-1}(I)$ is an open set in $tau_w$ containing $x$ so it contains $x_n$ for $n$ sufficiently large. Hence $f(x_n) in I$ for such $n$ proving that $f(x_n) to f(x)$. Conversely, suppose $f(x_n) to f(x)$ for every $f in X'$. Let $V$ be a neighborhood of $x$ in $tau_w$. Then there exist $f_1,f_2,...f_k in X'$ and an open intervals $I_1,I_2,...,I_k$ such that $x in cap_j f_j^{-1}(I_j) subset V$. Hence $f_j(x) in I_j$ for each $j leq k$ which implies $f_j(x_n) in I_j$ for $n$ sufficiently large. It follows that $x_n in f_j^{-1}(I_j) subset V$ for $n$ sufficiently large. Hence $x_n to x$ in $tau_w$.
answered Jan 12 at 23:23
Kavi Rama MurthyKavi Rama Murthy
59.5k42161
$endgroup$
In 1) 'is the topology ...' is is wrong. There are many topologies with this property and weak topology is the smallest topology for which $f^{-1}(I) in tau_w$ for every open interval $I$. It is the topology for which sets of the form $f^{-1}(I)$ form a subbase. (Every open set in $tau_w$ is a union of finite intersections of sets of this type).
Suppose $x_n to x$ in $tau_w$ and $fin X'$. Let $I$ be an open interval containing $f(x)$. Then $f^{-1}(I)$ is an open set in $tau_w$ containing $x$ so it contains $x_n$ for $n$ sufficiently large. Hence $f(x_n) in I$ for such $n$ proving that $f(x_n) to f(x)$. Conversely, suppose $f(x_n) to f(x)$ for every $f in X'$. Let $V$ be a neighborhood of $x$ in $tau_w$. Then there exist $f_1,f_2,...f_k in X'$ and an open intervals $I_1,I_2,...,I_k$ such that $x in cap_j f_j^{-1}(I_j) subset V$. Hence $f_j(x) in I_j$ for each $j leq k$ which implies $f_j(x_n) in I_j$ for $n$ sufficiently large. It follows that $x_n in f_j^{-1}(I_j) subset V$ for $n$ sufficiently large. Hence $x_n to x$ in $tau_w$.
answered Jan 12 at 23:23
Kavi Rama MurthyKavi Rama Murthy
59.5k42161
edited Jan 14 at 23:24
answered Jan 12 at 23:23
Kavi Rama MurthyKavi Rama Murthy
59.5k42161
answered Jan 12 at 23:23
Kavi Rama MurthyKavi Rama Murthy
59.5k42161
answered Jan 12 at 23:23
Kavi Rama MurthyKavi Rama Murthy
59.5k42161
59.5k42161
$begingroup$
Why is the set of the form $f^{-1}(I)$ a base? Can we only say that it is just a subbasis as the other person below said?
$endgroup$
– mmnn
Jan 14 at 15:42
$begingroup$
@mmnn My answer was faulty. I have corrected the answer.
$endgroup$
– Kavi Rama Murthy
Jan 14 at 23:26
add a comment |
$begingroup$
Why is the set of the form $f^{-1}(I)$ a base? Can we only say that it is just a subbasis as the other person below said?
$endgroup$
– mmnn
Jan 14 at 15:42
$begingroup$
@mmnn My answer was faulty. I have corrected the answer.
$endgroup$
– Kavi Rama Murthy
Jan 14 at 23:26
$begingroup$
Why is the set of the form $f^{-1}(I)$ a base? Can we only say that it is just a subbasis as the other person below said?
$endgroup$
– mmnn
Jan 14 at 15:42
$begingroup$
Why is the set of the form $f^{-1}(I)$ a base? Can we only say that it is just a subbasis as the other person below said?
$endgroup$
– mmnn
Jan 14 at 15:42
$begingroup$
@mmnn My answer was faulty. I have corrected the answer.
$endgroup$
– Kavi Rama Murthy
Jan 14 at 23:26
$begingroup$
@mmnn My answer was faulty. I have corrected the answer.
$endgroup$
– Kavi Rama Murthy
Jan 14 at 23:26
add a comment |
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.
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%2f3071453%2fquestion-about-weak-topology-on-normed-space%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
$begingroup$
$X'$ is usually called $X^*$ ("$X$-star").
$endgroup$
– DanielWainfleet
Jan 13 at 2:59