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# i have a math problem ><

are closures and interiors of connected sets always connected ?(look at subset of R^2) please step by step ! proof everything in detail! you may give an example for this case.

### 1 Answer

- SamLv 66 years agoFavorite Answer
The interiors of connected sets are not always connected.

counter example:

S={(x,y)|(x+1)^2+y^2 <=1} U {(x,y)|(x-1)^2+y^2 <=1}

The interiors of connected set S is

T={(x,y)|(x+1)^2+y^2 <1} U {(x,y)|(x-1)^2+y^2 <1}

which is disconnected.

2014-09-27 04:24:34 補充：

The closures of connected sets are always connected.

Hint :

S is connected < = > any continuous function f: S -> {0,1} is constant.

2014-09-27 14:54:35 補充：

The interiors of connected sets are not always connected.

counter example:

S={(x,y)|(x+1)^2+y^2 <=1} U {(x,y)|(x-1)^2+y^2 <=1}

The interiors of connected set S is

T={(x,y)|(x+1)^2+y^2 <1} U {(x,y)|(x-1)^2+y^2 <1}

which is disconnected. The set T is the union of two disjoint opendisks,

one is the open disk of center (-1,0) and radius 1,

and the other is the open disk of center (1,0) and radius 1.

So T is disconnected.********The closures V ofconnected sets U are always connected. hint : S is connected< = > any continuous function f: S -> {0,1} is constant….(1)[pf]For any continuousfunction f : V - > {0,1}, if we can prove that f is constant(or single value), then by (1) V isconnected.Consideringf : U - > {0,1}, f is continuous since f : V - > {0,1} is continuous. For U is connected, by(1) f is constant. Without loss of generalitywe assume f(u)=1 for all u in U. …(2)Let v be any pointin V, we have v=lim(n - > 00) u_n, where u_n are in U. f(v)=f(limu_n){By the continuity of f, we can change f and lim}=limf(u_n){by (2) f(u_n)=1}=lim1=1Impliesf(v)=1 for all v in V.By(1) V is connected.[endof proof]

2014-09-27 14:59:22 補充：

hint : S is connected< = > any continuous function f: S -> {0,1} is constant….(1)

ref: WIKI

http://en.wikipedia.org/wiki/Connected_space:

Formal definition, items (1) and (6).