the continuous image of a compact space is compact

Consider f:XY a continuousMathworldPlanetmathPlanetmath and surjective function and X a compact set. We will prove that Y is also a compact set.

Let {Va} be an open covering of Y. By the continuity of f the pre-image by f of any open subset ( of Y will also be an open subset of X. So we have an open covering {Ua} of X where Ua=f-1(Va).

To see this remember, since the continuity of f implies that each Ua is open, all we need to prove is that aUaX. Consider xX, we know that since {Va} is a covering of Y that there exists i such that f(x)Vi but then by construction xUi and {Ua} is indeed an open covering of X.

Since X is compact we can consider a finite setMathworldPlanetmath of indices {ai} such that {Uai} is a finite open covering of X, but then {Vai} will be a finite open covering of Y and it will thus be a compact set.

To see that {Vai} is a covering of Y consider yY. By the surjectivity of f there must exist (at least) one xX such that f(x)=y and since {Uai} is a finite covering of X, there exists k such that xUak. But then since f(Uak)=Vak, we must have that yVak and {Vai} is indeed a finite open covering of Y.

Title the continuous image of a compact space is compact
Canonical name TheContinuousImageOfACompactSpaceIsCompact
Date of creation 2013-03-22 15:52:48
Last modified on 2013-03-22 15:52:48
Owner cvalente (11260)
Last modified by cvalente (11260)
Numerical id 7
Author cvalente (11260)
Entry type Proof
Classification msc 54D30
Related topic CompactnessIsPreservedUnderAContinuousMap