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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.

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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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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 4Let $U,V$ be a neighbourhoods of zero, then1) 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 $\{V_{x_i}:i\in\{1,\ldots,n\}\}$ and $V \neq \emptyset$ because $0\in V_{x_i}$ for all $i$. 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.

$$

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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