Logical trap in R topology We know that in metric spaces, Bolzano-Weierstrass (BW) (each infinite set owns a cluster point) and Borel-Lebesgue (BL) properties are equivalent, i.e. compactness and countably compactness are equivalent (and also sequential compactness).
In the set of real numbers $\mathbb{R}$, the BW theorem is often written : "Any bounded infinite set has a cluster point". So according to the equivalence, we can say that in $\mathbb{R}$, "any bouded set verifies (BL)", which is not true because we know that only bounded and closed sets do...
Can you help me to find where I am wrong among this...?
EDIT : to be more precise, here is the wrong reasoning: 
"In $\mathbb{R}$, the collection of all infinite bounded sets verifies the (BW) property; since this property is equivalent to the (BL) property and since this one means compactness, all infinite bounded sets is compact."
 A: A sequence of points in a bounded but not closed subset will have a cluster point in $\mathbb R$, but the cluster point may fail to be a member of the subset -- so it is not necessarily sequentially compact.
A: OK, thanks to Henning Makholm, I realized my mistake. 
The definition of a cluster point refers to a subset $E$ of a topological space $X$, whereas the notion of compactness refers to the topology in itself, i.e. $E$ is a compact set ("of $X$") if and only if it is compact as a topological space for the topology of subspace. Hence, for compactness, the (BW) property must be studied from the viewpoint of the topological space, and not from the viewpoint of $E$ being a subset of $X$.
For example, in $X=\mathbb{R}$, it is true that any infinite subset of $E=]0,1]$ has a cluster point in $X$, even the infinite subset $A=\{\frac{1}{n} / n\in\mathbb{N^*}\}$ which has $0$ as cluster point in $X$. So $E$ verifies in $X$ the property that lies in (BW), but $E$ is not compact (or "is not a compact subset of $X$") because as a topological space, $E$ doesn't fulfil (BW) since the infinite set $A$ has no cluster point in $E$.
