Given any real number $x$, does there exist a smallest real number $y$ that is strictly greater than $x$?
More precisely:
Given $x \in \mathbb R$ does there exist $y \in \mathbb R$, such that $x<y$ and $\forall z\in \mathbb R((x<z \land z\not =y )\Rightarrow y \leq z)$.
A related question (regarding a special case) would be whether $\bigcap \{(0,a]: a \le 1\} = \emptyset \,$? which seems to be true given that $\emptyset = (0,0] \in \{(0,a]: a \le 1\}$.
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No, it does not (you can construct a sequence $y_n:=x+\frac1n$, decreasing to $x$...)
And yes $\bigcap_{a>0} (0,a] =\emptyset$
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Suppose on the contrary that such number exists, note that we have $$x<\frac{x+y}{2}<y,$$
hence a contradiction.
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Suppose that $y$ exists, then what about $\frac{x+y}{2}$ ?
EDIT : just saw that someone posted the same answer 50 seconds before me.
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As others have written, the answer is no.
However, there is something called the hyperreals which introduces something similar to this concept, so I'd like to share it.
I won't go over the construction, because it's fairly complicated. The set of hyperreals is denoted by $\Bbb R^*$ (or sometimes $^*\Bbb R$). The reals are a subset of the hyperreals; $\Bbb R\subset\Bbb R^*$. The set $\Bbb R^*$ has many convenient properties that make it useful. The relevant piece for us is, you can think of the hyperreals as the reals with "infinitesimals added in".
The hyperreals don't quite satisfy your criterion; there is no smallest positive hyperreal smaller than every other positive hyperreal. However, there are positive hyperreals smaller than every positive real. These numbers are called "infinitesimals" by people working with the hyperreals.
If $\epsilon$ is an infinitesimal hyperreal, so is $\frac{\epsilon}2$. So you do have things smaller than an infinitesimal. But you don't have any real thing smaller than an infinitesimal; anything between $0$ and $\epsilon$ is not real.
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As others said before me the answer is no. The reason is that if you fix two real numbers $a$ and $b$ with $a<b$ there are infinite other numbers in the interval $(a,b)$. You can use an iterative procedure to understand it: call $b_1=(a+b)/2 \Rightarrow a<b_1<b$, call $b_2=(a+b_1)/2 \Rightarrow a<b_2<b_1$, ..., call $b_n=(a+b_{n-1})/2 \Rightarrow a<b_n<b_{n-1}$ and so on. You can continue the procedure to infinity, with any value of $n \in \mathbb {N}$. This demontrate that you can always produce a real number as close as you want to $a$ but stricly greater than $a$. $a<b_n$, even if $n \to \infty$.
By the way, the same thing is valid also for rational numbers: $\forall \, a, b \in \mathbb {Q} \, , \, \exists \, b_n \in \mathbb {Q}$ with $a<b_n, \forall n \in \mathbb {N}$.
No, the real numbers don't have this property.
Suppose there is a smallest real number $x$ strictly greater than $0$; then consider $\tfrac{1}{2}x$ which satisfies: $$0 < \tfrac{1}{2}x < x$$
For real numbers, there is no such thing as a "successor": between two different real numbers $x$ and $y$, you always have their mean $\tfrac{x+y}{2}$.