[Math] Prove $b$ union its limit points is closed

general-topology

Let $b$ be a set of real numbers and $L$ the set of its limit points.

Prove, directly from the definition, that $b \cup L$ is closed.

I am having trouble to prove this. Could anyone help me with that? Thanks

Best Answer

Your definition is that a subset of $\Bbb R$ is closed if it contains all of its limit points, so in order to show that $A\cup L$ is closed, you must show that if $x$ is a limit point of $A\cup L$, then $x\in A\cup L$.

Suppose that $x$ is a limit point of $A\cup L$. Then for each $\epsilon>0$, $(x-\epsilon,x+\epsilon)$ contains a point of $A\cup L$ different from $x$. If every one of those open intervals contains a point of $A$ different from $x$, then $x$ is a limit point of $A$, so $x\in L\subseteq A\cup L$, and we’re done. Suppose, hoping to get a contradiction, that $x$ is not a limit point of $A$. Then there is an $\epsilon>0$ such that $(x-\epsilon,x+\epsilon)$ is either empty or just $\{x\}$. On the other hand, $x$ is a limit point of $A\cup L$, so $(x-\epsilon,x+\epsilon)$ contains some $y\in A\cup L$ with $y\ne x$. Clearly this means that $y\in L\setminus A$. Now let $\delta=\min\{y-(x-\epsilon),(x+\epsilon)-y\}$.

  1. Show that $(y-\delta,y+\delta)\subseteq(x-\epsilon,x+\epsilon)$.
  2. Explain why $(y-\delta,y+\delta)\cap A\ne\varnothing$.
  3. Explain why this is a contradiction.
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