Let $X \neq \lbrace 0 \rbrace$ be a reflexive space, and let $f \in X'$. Show that there exists $x \in X$ with $\|x\|=1$ and $f(x)=\|f\|$.  The definition of a reflexive space: $X$ is called reflexive if $J_X(X)=X''$ where $J_X: X \to X''$ defined by $J_Xx(f)=f(x)$ with $x \in X$ and $f \in X'$.  How to find this such $x$?
123147
Let $X\ne\{0\}$ be a reflexive space and let $f\in X^*$, where $X^*$ is the dual of $X$. I want to know: in general, does there exist an $x\in X$ with $\|x\|=1$, and $f(x)=\|f\|$, where $\|f\|$ is defined as $\sup\{|f(x)|:x\in X,\|x\|=1\}$?  I know this is true for $\mathbb{R}^n$ with the norm from the standard inner product, but I'm wondering if it is true in general.
53961
Let $X,Y$ be Banach spaces and $A\in\mathcal L(X,Y)$ . The task is to prove the following:  $A$ is compact if and only if the image of the closed unit ball in $X$ is compact in $Y$.  I have proven this when $X$ is a reflexive space.  Proof. Let $X$ be a reflexive space, $\bar B$ the closed unit ball in $X$, and $A$ a compact operator. Let further $y_n=Ax_n$ be a sequence in $A(\bar B)$.  In reflexive spaces $\bar B$ is weakly compact, so there exists a subsequence $x_{n_j} \to x$ weakly.  Because $A$ is compact, $Ax_{n_j}\to Ax$ strongly.   On the other side, $A(\bar B)$ is relatively compact, so there exists $z_k=Ax_{n_{j_k}}$ that converges strongly to $y\in Y$.  But $z_k\to Ax$ strongly. So by unicity of the limit $y=Ax$ and the image is compact.  It's easily proved that if the image is compact, the operator is also compact.  But I don't know what to do in case of nonreflexive spaces. Is there any counterexample or proof in such case?
58067
I have the following problem:      Let $(X,\tau)$ be a topological space and $(Y,\tau_y)$ be its    subspace, that is $Y\subset X$ and $\tau_y=\{\;A\;\cap\;Y\mid  A\in\tau\;\}$. Let also $F\subset Y$ and $H\subset X$, where $H$ is    closed in $X$. Show that $F=Y\;\cap\;H$ if and only if $F$ is closed   in $Y$.   I solved this problem by assuming that if $F$ is closed then $F=Y\;\cap\;H$, where $H$ is closed in $X$, but I got stuck proving this in the other direction, that is showing that if $F=Y\;\cap\;H$ then $F$ is closed in $Y$.  I tried to do something like this:   Assumption: $F=Y\;\cap\;H$, where $H$ is closed in $X$. If $H$ is closed in $X$ then $X\;\setminus\;H$ is open in $X$ and then $X\;\setminus\;H \in \tau$ which means that $B = (X\;\setminus\;H)\;\cap\;Y\in \tau_y$.  How should I proceed now? Should I investigate the set $B$ now more or what?
16374
For all $x,y\in\Bbb{R}$ define that $x\equiv y$ if $x^2=y^2$ . Then $\equiv$ is an equivalence relation on $\Bbb{R}$ , there are infinitely many equivalence classes, one of them consists of one element and the rest consist of two elements.  Solution  True. To show that $\equiv$ is reflexive we need to show that $\forall x\in\mathbb{R} :x\equiv x$. Let $x\in\mathbb{R} , x\equiv x \mbox{ if } x^{2}=x^{2}$, which is obvious.  $[x] ={y\in\mathbb{R} |x\equiv y} =[0]={y\in\mathbb{R}|0\equiv y}={0}$. Hence $y^{2} =0^{2} =0$ which implies that $y=0$.  Can anyone please give me correct answer to this question.
79259
I don't understand the task from (b). Is it equivalent to : for every linear functional $f(x_{1}, ..., x_{n})=c_{1}x_{1}+...+c_{n}x_{n}$ on $F^{n}$ which satisfy $c_{1}+...+c_{n}=0$ there exists exactly one and unique functional that belongs to the dual space of $W$?  If it is so, could someone please give me a hint?
31659
I have a doubt. I need help.  Can the basic axioms of equality be presented as "open axioms"?  I) (reflexivity) $\qquad x = x$              II)  (Substitutivity) $\qquad (x = y) \to \big(F (x, x) \to F (x, y)\big)$
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Here's a solution (to the exercise "prove that if $X$ is a space with more than one element, and is normal and connected then $X$ is uncountable"):   by Urysohn's lemma, given $A$ and $B$, closed and disjoint in $X$, there exists a continuous function from $X$ into $[0,1]$ such that $f(A)={0}, f(B)={1}$.   If $X$ were countable, so it would be $f(X)\subset [0,1]$; choose $r\in (0,1)\setminus f(X)$, then $X=f^{-1}([0,r))\cup f^{-1}((r,1])$, so $X$ is disconnected.  But the full power of the lemma wasn't used, all that was used is that there exists a continuous function $X\longrightarrow [0,1]$, so the condition of being normal seems too much.  So my question is, what are the weakest conditions on $X$ for the existence of such continuous functions into $[0,1]$ (or $\mathbb{R}$, for that matter)?
29943
This is what I have so far:   First, we will attempt to show $A-(B-C) \subseteq (A-C) - C.$     Let $x \in A -(B-C)$.      Then $x\in A$ and $x \notin (B-C)$. By DeMorgan's we have that  $$x \notin (B-C) = x \notin B \vee x \in C.$$     We have that $x \in A \wedge (x \notin B \vee x \in C)$.      Then we have that $(x \in A \wedge x \notin B) \vee (x \in A \wedge x \in C)$.  So by definition of $\cup$ and $\cap$, we have $x \in (A-B) \cup (x\in (A \cap C))$. I'm kind of stuck here.  I can see a contradiction here in that LHS says $x \in C$ but the RHS says $x \notin C$.   Can anyone tell me what to do from here? I'm not quite sure how to formally state that the statement is false. I feel like I'm not being complete here.
50232
Proof:- From (c) $X\in \mathscr T$. $X\subset cl(X)[\because by (a)]$ and also we have $cl(X)\subset X$. Hence $cl(X)=X$ and $\emptyset \in \mathscr T$.   Let $\{U_i\}_{i=1}^n$ be a collection of open sets in $\mathscr T$. For each $i$ there exists $C_i$:$cl(C_i)=C_i$ and $U_i=X\setminus C_i$ then $\bigcap_{i=1}^{n}U_i=X\setminus \bigcup_{i=1}^n C_i$. We have $cl(C_i)=C_i,i=1,2,3,..,n$. By (d), we have $\bigcap_{i=1}^{n}U_i \in \mathscr T$. How do I prove arbitrary union axiom? If there is an axiom $\bigcap_{\alpha \in \Lambda}cl(A_\alpha)=cl(\bigcap_{\alpha \in \Lambda}A_\alpha)$. I could prove. Can you please help me?
11551
I am trying to solve the following problem.     Let $X$ be a compact Hausdorff space and let $f:X\to X$ be continuous.   Show that there exists a non-empty set $A\subset X$ such that   $f(A)=A$.   There is a hint to define $A=\cap_{n\ge 0} A_n$ with $A_{n+1}=f(A_n)$ and $A_0=X$.  Then $x\in A_{n+1}$ implies that $f^{-1}(x)\cap A_n\ne\emptyset$. Right? So if $f^{-1}(x)\cap A_n=\emptyset$ then $x\notin A_{n+1}$. Is this correct?  First I need to show that for $A$ defined in the hint it holds $f(A)=A$. For this to happen it is enough to have $f(A)=\cap_{n\ge 0} f(A_n)$ Because then $f(A)=\cap_{n\ge 1} A_n=\cap_{n\ge 1} A_n\cap X=A$. But got stuck proving the equality.  What I did for that is to assume that there exists $x\in A$. This means that $\exists y_n\in A_n$ such that $f(y_n)=x$ for all $n$. Is this enough to deduct that $f(x)\in A$?  Then I need to show that $A$ is not empty. For this I think it is enough to show that $A_{n+1}\subset A_n$, because $A_n$ are compact and then their intersection is not empty. However I am stuck there too. I tried by contradiction assuming $\exists x\notin A_n$ with $f(x)\in A_{n}$, but I cannot work from there.  Initially I thought of something else. Choose $x_n\in A_n$ and create the sequence $\{x_n\}$. This has a convergent subsequence $\{x_{n_k}\}$. Then I create a sequence $\{x'_m\}$ by defining $x'_m=x_{n_k}$ with $n_k=\max\{j\in\{n_k\}|j\le m\}$. So basically i repeat the terms of the subsequence to get a convergent sequence. However this again requires that $A_{n+1}\subset A_n$ so I have $x'_m\in A_m$. Then I think that the limit $x'_m\to x$ is in $A$ but I was not able to show that also.  Thanks in advance.
891
Does $f(f(f(x)))=x \implies  f(x)=x$? Is it necessary for it to follow? How do we prove this? Do we have to substitute some special $x$? Or is it some other consideration of its properties?  Edit: does it follow in the special case when f is surjective? Or would we wind up with the same thing?
114844
Let $f:X\to X $ be continuous. Show that if $X=[0,1] $, $f $ has a   fixed point(i.e. there exists  $x$ such that $f (x)=x$). What happens   if $X $ equals  $[0,1) $ or $(0,1) $?   First part of the question is an immediate consequence of intermediate value theorem(for a proof, see ). I think $f (x)=x^2$ is a counterexample when $X=(0,1) $, since $x^2\lt x $ for all $x\in (0,1) $. But to be honest, I don't understand what causes breakdown of the fixed point theorem on $(0,1)$, since IVT only requires connectedness of domain. Is this related to the non-compactness of $(0,1)$?  Also I can't think of any counterexample for the case $X=[0,1)$(assuming fixed point is somehow related to compactness).  Any help is appreciated.   Thank you.
121330
So the idea that I am having is, saying that we let f be surjective, and we would prove that $f^{-1}$ is injective. What is confusing me is how I would word my proof and how I would represent it. I am starting it off with Let X and Y be non-empty sets, and let $f: X\rightarrow Y$ be a function. Now I'm stuck.
24705
Please give me feedback on my answer to this question. Question:  For all $x,y\in R$ define that $x\equiv y$ if $x^{2}=y^{2}$. Then $\equiv$ is an equivalence relation on $R$, there are infinitely many equivalence classes, one of them consists of one element and the rest consist of two elements. Answer: False, since;  $x\equiv y$   if $x^{2}=y^{2}$  on $R$  To show $\equiv$   is reflexive, we need to show that ,  $\forall x\in R:X\equiv x.$  Let $x\in R$,   $X\equiv x$   if $X^{2}=x^{2}$  , which is obvious.  $[x]\triangleq:\{y\in R/x\equiv y\}$  $[0]=[y\in R/0\equiv y\},$  when $y=0$,  Then $y^{2}=0^{2}=0$  Thus; there are not many infinity many equivalence classes.
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I am struggling with an exercise in Topology. The exercise goes:  Let $Y$ be Hausdorff, and $X\subseteq Y$ an open subspace which is locally compact. Let $X^{+}$ denote the one-point compactification of $X$, and let $\infty$ denote the unique point in $X^{+} \backslash X .$ Define $f: Y \longrightarrow X^{+}$ by $$ f(y)=\left\{\begin{array}{ll} y &amp; \text { if } y \in X \\ \infty &amp; \text { else } \end{array}\right. $$ Show that $f$ is continuous.  I have a lot of definitions and lemmas about continuity, but none of them seems to fit with this problem. I am looking for a hint on how to get started with the proof. Thanks in advance.
42395
If $f$ is continuous from $ [0,1]$ to itself and $f(0)=0$, $f(1)=1$ and $f(f(x))=x$ then show that $f(x)=x$ for every $ x $ in the domain of function.  I'd like to prove this statement, my idea goes like below:  Suppose that the converse of the statement holds. then without loss of generality, assume that there exists some $c$ in $[0,1]$ such that $d=f(c)&gt;c$. By symmetry, $f(d)=f(f(c))&lt;f(c)=d$. And $f$ is bijective because $f$ has an inverse function in this domain. So $f$ should be strictly increasing function. But on $[c,d]$, $f$ is decreasing. This is a contradiction.  So it should be that $f(x)=x$ on $[0,1]$.  Is this argument right? And I'd like to know another argument which prove this statement. Thank you for your answer in advance :)
88050
A collection $\{V_{\alpha}\}$ of open subsets of $X$ is said to be a base for $X$ if the following is true: For every $x\in X$ and every open set $G\subset X$ such that $x\in G$, we have $x \in V_\alpha \subset G$ for some $\alpha$. In other words, every open set in $X$ is the union of а subcollection of $\{V_{\alpha}\}$.   Prove that every separable metric space has a countable base.   Proof: Let $(X,d)$ be a metric space and $D=\{d_j\}_{j\in \mathbb{N}}$ be the countable and dense subset of $X$. Let $U$ is an open set in $X$ then for $x\in U$ $\exists \varepsilon_x&gt;0$ such that $N_{\varepsilon_x}(x)\subset U$. Then $\exists d_j\in D$ such that $d(x, d_j)&lt;{\varepsilon_x}/{4}$ $\Rightarrow$ $d_j\in N_{{\varepsilon_x}/{4}}(x)\subset U$.   It's easy to check that $N_{{\varepsilon_x}/{4}}(d_j)\subset N_{{\varepsilon_x}}(x)$. Indeed, if $z\in N_{{\varepsilon_x}/{4}}(d_j)$ then $d(z,d_j)&lt;\varepsilon_x/4$ and since $d(x, d_j)&lt;{\varepsilon_x}/{4}$ then by triangle inequality: $$d(z,x)\leqslant d(z,d_j)+d(x, d_j)&lt;{\varepsilon_x}/{4}+{\varepsilon_x}/{4}&lt;{\varepsilon_x}.$$  Hence $z\in N_{{\varepsilon_x}}(x)$ and inclusion $N_{{\varepsilon_x}/{4}}(d_j)\subset N_{{\varepsilon_x}}(x)$ holds.  By the way $N_{r}(d_j)\subset N_{{\varepsilon_x}}(x)\subset U$ for $r\in \mathbb{Q}$ and $r&lt;{\varepsilon_x}/{4}$. Thus for every $x\in U$ we associate an open ball with center at some point $d_j$ of $D$ and rational radius $r$ such that $x\in N_{r}(d_j)\subset N_{{\varepsilon_x}}(x)\subset U$.  Thus $$U=\bigcup\limits_{d_j\in U} N_{r_j}(d_j).$$ This set equality can be verified very easy. Inclusion $\subseteq$ I described in above paragraph. Converse inclusion is not difficult: if $z\in \bigcup\limits_{d_j\in U} N_{r_j}(d_j)$ then $z\in N_{r_j}(d_j)$ for some $d_j\in U$. But these balls  arranged so that $N_{r_j}(d_j)\subset U$ then $z\in U$.  Hence collection of balls $\{N_{r}(d): r\in \mathbb{Q}, d\in D\}$ is a countable base.  Is my proof correct? Can anyone check it please? Sorry if this topic is repeated but I would like to know is my proof correct.