Is that true that 2-sphere in $\Bbb R^4$ is like circle in $\Bbb R^3$ (having hole) Circle in $\Bbb R^2$ is a closed curve but viewing it in $\Bbb R^3$ we know that one can enter from one side and go into hole of circle then one can leave the hole of circle.

I think all spheres have such property but I can't justify this claim intuitively or in language of homology groups. For example $\Bbb S^2$ looks has no hole to enter into its hole but I think it is possible in $\Bbb R^4$. I don't know how to justify this? I think that the phrase "to enter into its hole" has no vague point because fortunately sphere has only one hole; i.e. $n$-dim hole at least on even-dimensional spheres.
Sorry for my non standard question.
 A: Interesting question!
As mentionned in the comments, what we mean by a hole is not obvious. Here the property I propose to use is: the sphere $S^{n-1}$, when put in $\mathbb{R}^{n+1}$, has a hole if we can show a path from its interior to its exterior that has an empty intersection with $S^{n-1}$.
Where "interior" and "exterior", or "inside" and "outside", refer to the situation of $S^{n-1}$ when in $\mathbb{R}^n$. Of course in $\mathbb{R}^{n+1}$ the sphere $S^{n-1}$ has no more any interior.
First the dim 3 ($\mathbb{R}^3$) usual case.
To prove that one circle $S^2$ has a hole, we can choose two circles as follows:

*

*$S^2_A$ has equations: $z=0, x^2+y^2=1$

*$S^2_B$ has equations: $y=0, (x-1)^2+z^2=1$
Properties:

*

*The two circles have empty intersection.

*$S^2_A$ contains the center $B=(1,0,0)$ of $S^2_B$.
$S^2_B$ contains the center $A=(0,0,0)$ of $S^2_A$.
These two centers are different points.

*Intersection of $S^2_B$ with plane $z=0$ containing $S^2_A$ has two points, one inside $S^2_A$ (that's the center $A$), and one outside $S^2_A$ (the point $(2,0,0)$, which is at distance $2$ from $A$ so outside of $S^2_A$). So there is a path between the interior of $S^2_A$ and its exterior, that does not intersect $S^2_A$.

Now the dim 4 case: we want to show that it is possible to have a sphere $S^3_A$ and a circle $S^2_B$ with similar properties as above. Let's have:

*

*$S^3_A$ has equations: $z=0, x^2+y^2+t^2=1$

*$S^2_B$ has equations: $y=0, t=0, (x-1)^2+z^2=1$
Properties:

*

*The two elements have empty intersection.

*$S^3_A$ contains the center $B=(1,0,0,0)$ of $S^2_B$.
$S^2_B$ contains the center $A=(0,0,0,0)$ of $S^3_A$.
These two centers are different points.

*Intersection of $S^2_B$ with plane $z=0$ containing $S^3_A$ has two points, one inside $S^2_A$ (that's the center $A$), and one outside $S^2_A$ (the point $(2,0,0,0)$, which is at distance $2$ from $A$ so outside of $S^3_A$). So there is a path between the interior of $S^3_A$ and its exterior, that does not intersect $S^3_A$.

This generalizes easily to any higher dimension $n$, as we can see by the case dim=4 being very similar to the case dim=3.
Intuitively, when we add one dimension to $S^{n-1}$ natural living environment, which is $\mathbb{R}^n$, i.e. when we put $S^{n-1}$ in $\mathbb{R}^{n+1}$, the sphere has no roof nor floor to protect it from  an invasion through the $n+1^{th}$ dimension.
The hole is then a one-dimension hole: it can be traversed by a line (infinite on one dimension).
When we add another dimension, i.e. put $S^{n-1}$ in $\mathbb{R}^{n+2}$, the hole becomes a two-dimension hole: it can be traversed by a plane (infinite on two dimensions). This is the same situation as a $S^0$ (= two distinct points) in $\mathbb{R}^3$.
And etc.
