general topology - Are closed subsets of a locally compact space, locally compact?

I know that there are open subsets of locally compact topological spaces that are not locally compact ($\mathbb{Q}$ in the Alexandroff's compactification).I wonder if any closed subset of a locally compact space is always locally compact.Definition. $(X,\tau)$ is locally compact if for each $x\in X$, there is a neighborhood of $x, A_x\in \tau:closure(A_x)$ is compact....Read more

functional analysis - what exactly is weak* topology?

I know that weak* topology is the weakest topology so that $Jx$ is continuous for $\forall x\in X$, where $J$ is the isometry from $X$ to $X''$. But what exactly is this topology? What is the open set in general look like?And moreover, I want to prove another topology induced by metric is exactly the weak* topology. How can I prove this topology is weaker than weak* topology, so that it is exactly the weak* topology?...Read more

Subbasis of a weak topology

Let $\{(X_\alpha,\mathscr{T}_\alpha):\alpha\in\Lambda\}$ be an indexed family of topological spaces, and for each $\alpha\in\Lambda$ let $f_\alpha:X\to X_\alpha$ be a function. Furthermore, let $\mathscr{T}$ be the weak topology on X induced by $\{f_\alpha:\alpha\in\Lambda\}$. Then $\mathscr{L}=\{f_\alpha^{-1}(U_\alpha):\alpha\in\Lambda; U_\alpha\in\mathscr{T}_\alpha\}$ is a subbasis for $\mathscr{T}$.I have tried this: Since $f_\alpha$ is continuous therefore $f_\alpha^{-1}(U_\alpha)$ is open then, intersections of elements of $\mathscr{L}$ ar...Read more

functional analysis - Weaker Than The Weak Topology?

The weak topology on a Banach Space $E$ is defined to have sub-base consisting of open balls of the form $B_\alpha(x,r) = \lbrace y \in E : \vert \alpha(x-y) \vert < r\rbrace $ for each $x \in E$ and $r > 0$ and $\alpha \in E^*$ . This defines a uniform structure which is also a topological group.Is there any reason we would want to define a similar topology, but only using a proper subset of the seminorms induced by dual space elements? That is the topology on $E$ with sub-base consisting of balls of the form $B_\beta(x,r) = \lbrace y \i...Read more

weak-* topology

What is the definition of weak-*topology ($\sigma(X^*,X)$)? What is the intuition standing behind its definition (apart from easier extraction of finite subcovers - Banach - Alaoglu theorem)?I start with weak topology. So let $X, Y_j$ be topological spaces and let $f_j\colon X\to Y_j$, $j\in J$. Consider a collection$$\mathcal{O}=\{\bigcap_{j=1}^k f_{j}^{-1}(O_j)\mbox{-open in}\,X,\, k\in\mathbb{N},\,O_j\, \mbox{open in}\, Y_j\}.\qquad (1)$$ The collection $\mathcal{O}$ is closed under finite intersections, so it is a base of a topology on $X$....Read more

functional analysis - weak topology has less open set than strong topology (in Banach spaces). Why?

Let $E$ a Banach spaces of infinite dimension. The weak topologie is the thickest that makes functional continuous. Let denote $\mathcal T_W$ the weak topology on $E$. 1) I call "Dual topological" the element of $E^*=\mathcal L(E,\mathbb R)$ that are continuous (wrt the strong topology that I denote $\mathcal T$, i.e. induced by the norm of $E$). I denote it $E'$. I know that $E'\subsetneq E^*$, i.e. there are linear functional that are not continuous.2) In the book "Analyse fonctionnelle : Théorie et application" of Haim Brezis it's written th...Read more

The topology generated by $\mathcal{A}$ is minimal

Let $X$ a set and $\mathcal{A}$ a base for the topology $\tau$. Consider $\{\tau_\alpha\}_{\alpha\in I}$ a family of all topologies over $X$ such that $\mathcal{A}\subset\tau_\alpha$ for any $\alpha\in I$. I want to show that $\bigcap_{\alpha\in I}\tau_\alpha=\tau$.Proof:$\tau\subset\bigcap_{\alpha\in I}\tau_\alpha$: Let $u\in\tau$ then, exists $V_\kappa\in\mathcal{A}$ such that $u=\bigcup V_\kappa$ and, as each $V_\kappa$ is a open set for each $\tau_\alpha$ $(\alpha\in I)$, we have that $u\in\tau_{\alpha}$ for $\alpha\in I$. So, we obtain $u\...Read more

general topology - Topological Proof that every Interval $I \subset \mathbb{R}$ is connected

First, the definition of a connected set: Definition: A topological space is connected if, and only if, it cannot be divided in two nonempty, open and disjoint subsets, or, similarly, if the empty set and the whole set are the only subsets that are open and closed at the same time.I don't understand some points in the following proof, that every interval $I \subset \mathbb{R}$ is connected.Suppose $I = A \cup B$ and $A \cap B = \emptyset$, $A$ and $B$ are both non-empty and open in the subspace-topology of $I \subset \mathbb{R}$. Choose $a\...Read more

general topology - Proposed proof of: If $A \subset B \subset \bar{A}$ and $A$ is connected, then $B$ is connected

Hi I just want to find out if the following is a acceptable proof for the proposition: "Consider metric space $(X,d)$. If $A \subset X$ is connected and $A \subset B \subset \bar{A}$ then $B$ is connected."Proof: Assume that $A \subset X$ is connected and $B$ is not connected. There exists disjoint open sets (in $(B,d)$) $V_{1}$ and $V_{2}$ such that $B = V_{1} \cup V_{2}$. Since $A \subset B$, it follows that $A \cap V_{1}$ and $A \cap V_{2}$ are open in $(A,d)$. Also then $A = (A \cap V_{1}) \cup (A \cap V_{2})$. Thus since $A $ is assumed co...Read more

general topology - Show that every open subset of a locally connected space is locally connected.

Suppose $X$ is a topological space that is locally conneted and let $O$ be an open subset of $X$. Then we want to show that $O$ is also locally connected.Let $p\in O$ chosen arbitrarily, then there exists a open set $U\subset O$ and a connected open set $V\subset X$, both of which containing $p$. Now their intersection $W=U\cap V$ is open and contains $p$, therefore $W$ is a neighborhood of $p$ and $W\subset U\subset O$. If $W$ is connected then we are done. Now, suppose that $W$ is not connected hence can be written as a disjoint union of two ...Read more

general topology - Continuous image of connected set is connected: Proof

In De La Fuente's Mathematical Methods and Models for Economists, the following is said: Let $f:X\to Y$ be a continuous mapping between two metric spaces. If C is a connected subset of $X$, then $f(C)$ is connected.The proof goes as in Rudin's Principles, and I cannot understand exactly what Rudin also does not explain: Suppose $f(C)$ is not connected. Then $f(C)=P\cup Q$, where $P$ and $Q$ are nonempty, separated subsets of $Y$, that is, $clP\cap Q = \emptyset$ and $P\cap clQ = \emptyset$ Let $$ A = C\cap f^{-1}(P) \\ B = C\cap f^{-1}(...Read more

general topology - About the proof: if $X_i$ is connected, then $\prod_{i\in I} X_i$ is connected.

Could someone explain to me why in the following proof $\bigcup Y_F=Y$ (is the bold text)? $\forall i\in I$, if $X_i$ is connected, then $\prod_{i\in I} X_i$ is connected.Proof. First prove that any finite product of connected spaces is connected. This can be done by induction on the number of spaces, and for two connected spaces $X$ and $Y$ we see that $X \times Y$ is connected, by observing that $X \times Y = (\cup_{x \in X} (\{x\} \times Y)) \cup X \times \{y_0\}$, where $y_0 \in Y$ is some fixed point. This is connected because every set $...Read more

general topology - In proving every open connected subset of $\mathbb{R}^n$ is path connected

I would like to prove that every open connected subset of $\mathbb{R}^n$ is path connected.Let us choose $E$ to be a such open connected subset, then given any point $p \in E$, we will define $F$ be the set of all points in E that can be joined to $p$ by a path in $E$.The idea is to show that $F = E$ by showing $F$ is clopen as well as $F$ is path connected.We choose $q \in F \subseteq E$, then since $E$ is open, we can find an open ball $D_x(q,\epsilon)$ in $E$.Here are my questions.My lecturer said $D_x(q,\epsilon)$ is path connected as any p...Read more

general topology - Compactness of a metric space

If a metric space $(X,d)$ is compact then for every equivalent metric $\sigma$, $(X,\sigma)$ is complete. This is because, for any cauchy sequence in $(X,\sigma)$ has a convergent subsequence due to fact $(X,\sigma)$ is a compact metric space, hence original sequence is convergent. My question is , does the converse also hold ? That is, let $(X,d)$ be a metric space such that for every equivalent metric $\sigma$, $(X,\sigma)$ is complete. Does this imply $(X,d)$ is compact?...Read more

general topology - Totally bounded, sequentially compact, complete, bounded, closed, equicontinuous $\Rightarrow$ compact?

Related; When $K$ is compact, if $S\subset C_b(K)$ is closed,bounded and equicontinuous, then $S$ is compact? (ZF)I just edited my whole question since i think it was a bit messy.Here is my question.Let $K$ be a separable compact metric space and $S\subset C(K,\mathbb{C})$.Let $S$ be closed,bounded,uniformly equicontinuous on $K$, sequentially compact, totally bounded and complete.Then is $S$ compact? (in ZF)Thank you in advance!...Read more