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For an arbitrary discrete space X, construct a compact topological space Y and a topological embedding Y.

I am think about construct $X={0,1}$ equipped with the discrete topology. Topology on Y={∅,{0,1},{1}} which is the Sierpinski space. Thus Y is immediately compact. I try to define the map as $f(∅)=∅， f(X)=X, f({0})=f({1})={1}$. Then the $ab:X->f(X)$ is homeomorphism. Would anyone help me to check if my idea is correct?

manzujizhe
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    Taking ${0,1}$ is far from arbitrary. Also $Y$ is the set ${0,1}$ or the set ${\varnothing,{1},{0,1}}$? The latter is not the Sierpinski space. Finally embeddings are injective by definition. – Asaf Karagila Oct 12 '13 at 10:04
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    There are standard constructions to do this. However, this looks like an exercise, and without context no-one can say what an appropriate answer is. – Carsten S Oct 12 '13 at 10:16

2 Answers2

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If $X$ is finite, $X$ is already compact, and you might as well let $Y=X$ with the identity map as your embedding. There’s no reason to add an extra point.

The exercise becomes interesting only when $X$ is infinite, and then adding another point is necessary. There is a way to do it by adding just one point: it’s called the Alexandrov one-point compactification of $X$. See if you can work it out from that little article; if you get stuck, feel free to leave a question for me.

Brian M. Scott
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Your idea is not correct, since $f(\emptyset)=\emptyset$, $f(X)=X$, $f(0)=0$, $f(1)=1$ does not define a function from $X$ to $Y$, let alone an injective one.

Carsten S
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