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I recently saw a proof of this using the fact that the star of a vertex $v$ of a simplicial complex is open. However, this does not hold if $st(v) = v$ where $st(v)$ means the star of $v$ (i.e. $v$ lies only in a zero-simplex).

Is there any reason as to why $ K = (V, V)$ where $V = \{ (0,0), (\frac{1}{n},0)$ | $n\in \mathbb{N}\}$ would not be a valid infinite simplicial complex? It is a closed subset of $\mathbb{R}^2$ as it contains all limit points. Further it is bounded and so is compact, which would seem to be a contradiction to the statement that no infinite simplicial complex is compact (equivalently that any compact simplicial complex is finite).

I have yet to cover cell complexes, so an argument without those would be very helpful.

Krishan Bhalla
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2 Answers2

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This answer addresses the question regarding the example $V$.

Simplicial complexes carry what is called the "weak topology", meaning that a subset is open if and only if its intersection with every simplex is an open subset of that simplex. From this it follows immediately that a $0$-dimensional simplicial complex has the discrete topology, because every subset intersected with every $0$-simplex is an open subset of that simplex. Your $0$-dimensional example fails to be discrete.

Lee Mosher
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Yes. A simplicial complex is compact iff it is finite. Proof by contradiction choose a point in every simplex. This gives you a discrete infinite set.

Your set is not a simplicial complex since countable simplicial complexes are discrete.

ThorbenK
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