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I came across this theorem and I am disappointed not being able to understand or to have intuition to understand it . I would be glad to get help . Theorem : If $K$ and $C$ are subset of topological vector space (TVS) $X$ , $K$ is compact , $C$ is closed and $K \cap C =\varnothing $ then $0$ has a nbhd $V$ such that $(K+V)\cap ( C+V) = \varnothing$

My first question is how to find a symmetric nbhd? I can't seem to understand the proof.

an example illustrating the continuity in TVS.

Concept behind local basis and how every other basis can be deduced from local basis.

I think one example would make me move forward.
Thank you for giving in your time.

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Given some neighborhood of zero $V$, you can construct symmetric neighborhood by considering $U=V\cap (-V)$. This open set containing $0$ such that $U=-U$ –  Norbert Jun 7 '12 at 9:47
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To gain geometric intuition, draw a picture in the Euclidean plane of $K+V$ and $C+V$ when $V$ is a small disk centered at the origin. –  Giuseppe Negro Jun 7 '12 at 9:49
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About your third point: if $\{U_a\}_{a\in A}$ is a basis of open neighborhood of $0$ then, for any $x\in V,$ a basis of open neighborhood of $x$ is $\{x+U_a\}_{a\in A},$ because the map $T_x:u\in V\to x+u\in V$ is a homeomorphism. –  Giuseppe Tortorella Jun 7 '12 at 9:51
    
@GiuseppeTortorella : If i have $K=\varnothing$ then why is $V+K=\varnothing$ ? –  Theorem Jun 7 '12 at 11:14
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@Ananda, If $V+K\neq\varnothing$, then there exist $v\in V$ and $k\in K$ such that $v+k\in V+K$. Since there exis some $k\in K$, then $K\neq\varnothing$. –  Norbert Jun 7 '12 at 11:21
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2 Answers

up vote 8 down vote accepted

I'll give extremely detailed explanation of theorem 1.10 from Rudin's Functional analysis. In this explanation I'll use a couple of standard results about topological vector spaces which you can skip if you already know them. Through all the text $X$ is a topological vector space.

Lemma 1. Let $a\in X$ and $\lambda\in \mathbb{C}\setminus\{0\}$, then the maps $$ T_{a}:X\to X:x\mapsto x+a\qquad M_\lambda:X\to X:x\mapsto\lambda x $$ are homeomorphisms.

Proof. Since addition and multiplication is continuous in topological vector spaces, then $T_a$ and $M_\lambda$ are continuous. By the same reasoning the maps $T_{-a}$ and $M_{\lambda^{-1}}$ are continuous too. Note that $$ T_a\circ T_{-a}=T_{-a}\circ T_{a}=\mathrm{Id}_X, $$ $$ M_{\lambda}\circ M_{\lambda^{-1}}=M_{\lambda^{-1}}\circ M_{\lambda}=\mathrm{Id}_X $$ Indeed, for all $x\in X$ we have $$ (T_a\circ T_{-a})(x)=(T_a(T_{-a}(x)))=T_a(x-a)=(x-a)+a=x $$ $$ (T_{-a}\circ T_{a})(x)=(T_{-a}(T_{a}(x)))=T_{-a}(x+a)=(x+a)-a=x $$ $$ (M_{\lambda}\circ M_{\lambda^{-1}})(x)=(M_{\lambda}(M_{\lambda^{-1}}(x))=M_\lambda(\lambda^{-1}x)=\lambda(\lambda^{-1}x)=x $$ $$ (M_{\lambda^{-1}}\circ M_{\lambda})(x)=(M_{\lambda^{-1}}(M_{\lambda}(x))=M_{\lambda^{-1}}(\lambda x)=\lambda^{-1}(\lambda x)=x $$ Since continuous maps $T_{a}$ and $M_\lambda$ have continuous inverses, then they are homeomorphisms.

Lemma 2. Let $x_0\in X$ and $V$ is a neighborhood of zero, then $x_0+V$ is a neighborhood of $x_0$.

Proof. By lemma 1 the map $T_{x_0}$ is a homeomorphism. Hence image of each open set under the map $T_{x_0}$ is open. In particular $T_{x_0}(V)=x_0+V$ is open. Since $x_0\in x_0+V$ and $x_0+V$ is open, then $x_0+V$ is a neighborhood of $x_0$.

Lemma 3. Let $V$ be neighborhood of zero and $F\subset X$ be some subset, then $F+V$ is an open set containing $F$.

Proof. Since $V$ is a neighborhood of zero, then from lemma 2 it follows that $$ F=\bigcup\limits_{x\in F}\{x\}\subset\bigcup\limits_{x\in F}(x+V)=F+V. $$ Thus $F\subset F+V$. By lemma 2 sets $x+V$ are open, then the set $F+V=\bigcup_{x\in F}(x+V)$ is open as union of open sets.

Lemma 4 Let $U,V$ be a neighbourhoods of zero, then

1) the sets $U\cap V$, $U\cup V$, $-U$ are neighborhoods of zero.

2) the set $U\cap(-U)$ is symmetric neighborhood of zero.

3) if $U$ is a symmetric neighbourhood of zero, then does $U+U$.

Proof. 1) Since $U$, $V$ are neighborhoods of zero, then $U$ and $V$ are open and $0\in U$, $0\in V$. Since $U$ and $V$ are open, then does $U\cap V$ and $U\cup V$. Since $0\in U$ and $0\in V$, then $0\in U\cap V$, $0\in U\cup V$. Hence $U\cap V$ and $U\cup V$ are neighborhoods of zero. Since $0\in U$, then $0=-0\in-U$. From lemma 1 we know that $-U=M_{-1}(U)$, i.e. $-U$ is an image of open set $U$ under homeomorphism $M_{-1}$, hence $-U$ is open. Since $-U$ is open and $0\in -U$, then $-U$ is a neighborhood of zero.

2) From previous paragraph it follows that $V\cap(-V)$ is a neighborhood of zero. Now we see that $$ -(V\cap(-V))=-(\{x:x\in V\}\cap\{-x:x\in V\})=\{-x:x\in V\}\cap\{-(-x):x\in V\}=\{-x:x\in V\}\cap\{x:x\in V\}=V\cap(-V), $$ hence $V\cap(-V)$ is a symmetric neighborhood of zero.

3) By lemma 3 $U+U$ is an open set. Since $U$ is a neighbourhood of zero $0\in U$, hence $0=0+0\in U+U$. Thus $U+U$ is a neighborhood of zero. Direct check shows $$ -(U+U)=-\{x+y:x\in U, y\in U\}=\{-x-y:x\in U, y\in U\}=\{\hat{x}+\hat{y}:\hat{x}\in -U, \hat{y}\in -U\}=\{\hat{x}+\hat{y}:\hat{x}\in U, \hat{y}\in U\}=U+U $$ So $U+U$ is a symmetric neighbourhood of zero.

Lemma 5. Let $W$ be a neighborhood of zero. Then there exist symmetric neighborhood of zero $V$ such that $V+V+V+V\subset W$.

Proof. Since $W$ is a neighborhood of zero, then $0\in W$ and $W$ is open, then from equality $0+0=0$ and continuity of addition in topological vector spaces it follows that there exist neighborhoods of zero $U_1$, $U_2$ such that $U_1+U_2\subset W$. By lemma 4 the set $U_0=U_1\cap U_2$ is a neighborhood of zero. By lemma 4 the set $U=U_0\cap(-U_0)$ is a symmetric neighborhood of zero. Now note that $$ U+U=\{x+y:x\in U, y\in U\}\subset \{x+y:x\in U_1,y\in U_2\}=U_1+U_2\subset W $$ Applying this resut to the neighborhood of zero $U$ we get symmetric neighborhood of zero $V$ such that $V+V\subset U$. Hence $$ V+V+V+V=\{x+y:x\in V+V,y\in V+V\}\subset\{x+y:x\in U,y\in U\}= U+U\subset W. $$

Theorem. Let $K\subset X$ is a compact, $C\subset X$ is closed and $K\cap C=\varnothing$, then there exist neighborhood of zero $V$ such that $(K+V)\cap(C+V)=\varnothing$.

Proof. Case 1. If $K=\varnothing$, then for arbitrary neighborhood of zero $V$ we have $K+V=\varnothing$. Indeed, if $K+V\neq\varnothing$, then there exist $x\in V$, $y\in K$ such that $x+y\in K+V$. Since there exist some $y\in K$, then $K\neq\varnothing$. Contradiction, hence $K+V=\varnothing$. Since $K+V=\varnothing$, then $(K+V)\cap(C+V)=\varnothing$.

Case 2. If $K\neq\varnothing$, then fix $x\in K$. Since $K\cap C=\varnothing$, then $x\notin C$ which is equivalent to $x\in X\setminus C$. Since $C$ is closed then $X\setminus C$ is open. Since $x\in X\setminus C$ and $X\setminus C$ is open then there exist open set $W_x\subset X\setminus C$ such that $x\in W_x$. Note that $X$ is a topological vector space, then by lemma 2 the set $W_x-x$ is also open. Since $x\in W_x$, then $0\in W_x-x$. Thus $W_x-x$ is an open set containing $0$, i.e. neighborhood of zero. Now by Lemma 5, there exist symmetric neighborhood of zero $V_x$ such that $V_x+V_x+V_x+V_x\subset W_x-x$. The set $V_x$ is a neighborhood of zero, so $0\in V_x$, hence $$ V_x+V_x+V_x=0+V_x+V_x+V_x\subset V_x+V_x+V_x+V_x\subset W_x-x. $$ Since we have inclusion $V_x+V_x+V_x\subset W_x-x$, then $x+V_x+V_x+V_x\subset W_x$. Recall that $W_x\subset X\setminus C$, hence $x+V_x+V_x+V_x\subset X\setminus C$. This is equivalent to $(x+V_x+V_x+V_x)\cap C=\varnothing$.

Assume that $(x+V_x+V_x)\cap(V_x+C)\neq\varnothing$, then there exist $z\in(x+V_x+V_x)\cap(V_x+C)$. Since $z\in(V_x+C)$, then we have $v\in V_x$ and $c\in C$ such that $z=v+c$. Then $c=z-v=z+(-v)$. Since $V_x$ is symmetric and $v\in V_x$, then $(-v)\in V_x$ This is the only place where we use that neighborhood is symmetric! Also note that $z\in(x+V_x+V_x)$, so $c=z+(-v)\in(x+V_x+V_x)+V_x=x+V_x+V_x+V_x$. But $c\in C$, hence $(x+V_x+V_x+V_x)\cap C\neq\varnothing$. Contradiction, so $(x+V_x+V_x)\cap(V_x+C)=\varnothing$. Thus for each $x\in K$ we constructed symmetric neighborhood of zero $V_x$ such that $$ (x+V_x+V_x)\cap(C+V_x)=\varnothing\tag{1} $$

Fix $x\in K$, then from Lemma 2 it follows that $x+V_x$ is a neighborhood of $x$. In particular $\{x\}\subset x+V_x$. Since $x\in K$ is arbitrary we conclude $K=\bigcup_{x\in K}\{x\}\subset \bigcup_{x\in K}(x+V_x)$. Thus we constructed a family $\{x+V_x:x\in K\}$ of open sets such that $K\subset \bigcup_{x\in K}(x+V_x)$ i. e. an open cover of $K$. Recall that $K$ is a compact, so there exist finite subcover, i.e. finite subfamily $\{x_i+V_{x_i}:i\in\{1,\ldots,n\}\}\subset\{x+V_x:x\in K\}$ such that $$ K\subset \bigcup_{i=1}^n(x_i+V_{x_i}). $$ Consider set $V=\bigcap_{i=1}^n V_{x_i}$ it is open as finite intersection of open sets $\{V_{x_i}:i\in\{1,\ldots,n\}\}$. Now we have inclusion $$ K+V=\{x+v:x\in K, v\in V\}\subset\left\{x+v:x\in\bigcup_{i=1}^n( x_i+V_{x_i}),v\in V\right\}\subset\bigcup_{i=1}^n\{x+v:x\in x_i+V_{x_i},v\in V\}=\bigcup_{i=1}^n\{x+v:x\in x_i+V_{x_i},v\in V\} $$ Since $V=\bigcap_{i=1}^n V_{x_i}\subset V_{x_i}$ for all $i\in\{1,\ldots,n\}$ then $$ K+V\subset \bigcup_{i=1}^n\{x+v:x\in x_i+V_{x_i},v\in V\}\subset $$ $$ \bigcup_{i=1}^n\{x+v:x\in x_i+V_{x_i},v\in V_{x_i}\}=\bigcup_{i=1}^n(x_i+V_{x_i}+V_{x_i})\tag{2} $$ Again since $V=\bigcap_{i=1}^n V_{x_i}\subset V_{x_i}$ for all $i\in\{1,\ldots,n\}$ then $$ C+V=\{x+v:x\in C, v\in V\}\subset\{x+v:x\in C,v\in V_{x_i}\}=C+V_{x_i}\tag{3} $$ From $(1)$ it follows that $(x_i+V_{x_i}+V_{x_i})\cap (C+V_{x_i})=\varnothing$ for all $i\in\{1,\ldots,n\}$, then using $(3)$ we conclude that $(x_i+V_{x_i}+V_{x_i})\cap (C+V)=\varnothing$ for all $i\in\{1,\ldots,n\}$. Taking union over $i\in\{1,\ldots,n\}$ we get $$ \bigcup\limits_{i=1}^n(x_i+V_{x_i}+V_{x_i})\cap (C+V)=\varnothing\tag{4} $$ Finally from $(2)$ and $(4)$ we conclude that $$ (K+V)\cap(C+V)=\varnothing. $$

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Thank you so much sir. You have bothered so much for me :) –  Theorem Jun 7 '12 at 18:42
    
Not at all :) 3 years ago I also put so much effort to understand proofs from that book! –  Norbert Jun 7 '12 at 18:44
    
:) i am learning from Rudin, any suggestions to approach the book ? –  Theorem Jun 7 '12 at 18:47
    
Prove every statement that you read from this book. Exercises are unnesessary, some of them are very difficult. Ask questions on MSE. If Rudin is your first book of functional analysis, then put it away for a while. Start with something more simple. –  Norbert Jun 7 '12 at 18:57
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I will assume some basic topology.

Since addition is continuous and $C$ is closed, the set $\{(x,k)\in X\times K\colon x+k\in C\}$ is closed. Since $K$ is compact, the image of that set under $X\times K\to X$ is also closed, but this is $A:=C-K$. Since $0\notin A$, $U:=X\setminus A$ is a neighborhood of $0$. Since subtraction is continuous, there is a neighborhood $V$ of $0$ such that $V-V\subset U$. This means that for any $v,w\in V$, $c\in C$, $k\in K$ we have $v-w\ne c-k$, or $k+v\ne c+w$. Hence $V$ has the desired property.

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Arrgh, did not notice that the question is old. –  Carsten Schultz Oct 23 '13 at 19:51
    
nice, but OP was not of the level of this proof –  Norbert Oct 23 '13 at 20:31
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