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"Can we take the time interval between the two events 1 and 2 as observed by the mentioned observer as proper time interval?"

No, if two events are simultaneous in any frame, the interval between them is a space-like one, not a time-like one. If you're not familiar with the idea of the three categories of intervals, time-like, space-like and light-like, see the "Spacetime intervals in flat space" section of the wikipedia spacetime article. If you know the coordinates of two events in some inertial coordinate system, you can figure out the coordinate differences $\Delta x$, $\Delta y$, $\Delta z$ and $\Delta t$ between the events (i.e. if the first event had x-coordinate $x_1$ and the second had $x_2$, then $\Delta x = x_2 - x_1$), and then the spacetime interval can be defined as $s^2 = \Delta x^2 + \Delta y^2 \Delta z^2 - c^2\Delta t^2$. If $s^2$ is negative the interval is time-like (and the absolute value of $s$ is the proper time between the two events along the world line of an inertial object that passes through both, so this would be the actual time measured on this object's clock between two events which both occurred right next to the object), if $s^2$ is positive the interval is space-like (and $s$ is then the proper distance along a straight line path joining the two events--see my answer here for a little more on what "proper distance" means physically), and if $s^2$ is zero the interval is light-like (meaning only something moving at the speed of light could have both events on its world line).

Also, note that this interval is frame-invariant, which means that if you switched to a different inertial coordinate systems, then even though the individual values of $\Delta x$, $\Delta y$, $\Delta z$ and $\Delta t$ could change, $s^2$ would be unchanged. This is analogous to the geometric case where if you describe two points in a 2D Euclidean plane using two Cartesian coordinate systems with differently-oriented $x$ and $y$ axes, the values of $\Delta x$ and $\Delta y$ may differ depending on the coordinate system, but the distance $d$ between the points, determined by the Pythagorean theorem to be $d^2 = \Delta x^2 + \Delta y^2$, would be the same in both coordinate systems.

Since the interval is frame-invariant, then if you calculate two events to have a space-like separation in the coordinates of one inertial frame, they will have the same space-like separation in all inertial frames, and thus it's meaningless to ask what the proper time between these two events is. And if the events are simultaneous in at least one frame, then in that frame $\Delta t = 0$, so naturally $s^2$ will be positive and the interval must be space-like.

"Can we take the time interval between the two events 1 and 2 as observed by the mentioned observer as proper time interval?"

No, if two events are simultaneous in any frame, the interval between them is a space-like one, not a time-like one. If you're not familiar with the idea of the three categories of intervals, time-like, space-like and light-like, see the "Spacetime intervals in flat space" section of the wikipedia spacetime article. If you know the coordinates of two events in some inertial coordinate system, you can figure out the coordinate differences $\Delta x$, $\Delta y$, $\Delta z$ and $\Delta t$ between the events (i.e. if the first event had x-coordinate $x_1$ and the second had $x_2$, then $\Delta x = x_2 - x_1$), and then the spacetime interval can be defined as $s^2 = \Delta x^2 + \Delta y^2 \Delta z^2 - c^2\Delta t^2$. If $s^2$ is negative the interval is time-like (and the absolute value of $s$ is the proper time between the two events along the world line of an inertial object that passes through both, so this would be the actual time measured on this object's clock between two events which both occurred right next to the object), if $s^2$ is positive the interval is space-like (and $s$ is then the proper distance along a straight line path joining the two events--see my answer here for a little more on what "proper distance" means physically), and if $s^2$ is zero the interval is light-like (meaning only something moving at the speed of light could have both events on its world line).

Also, note that this interval is frame-invariant, which means that if you switched to a different inertial coordinate systems, then even though the individual values of $\Delta x$, $\Delta y$, $\Delta z$ and $\Delta t$ could change, $s^2$ would be unchanged. This is analogous to the geometric case where if you describe two points in a 2D Euclidean plane using two Cartesian coordinate systems with differently-oriented $x$ and $y$ axes, the values of $\Delta x$ and $\Delta y$ may differ depending on the coordinate system, but the distance $d$ between the points, determined by the Pythagorean theorem to be $d^2 = \Delta x^2 + \Delta y^2$, would be the same in both coordinate systems.

Since the interval is frame-invariant, then if you calculate two events to have a space-like separation in the coordinates of one inertial frame, they will have the same space-like separation in all inertial frames, and thus it's meaningless to ask what the proper time between these two events is. And if the events are simultaneous in at least one frame, then in that frame $\Delta t = 0$, so naturally $s^2$ will be positive and the interval must be space-like.

"Can we take the time interval between the two events 1 and 2 as observed by the mentioned observer as proper time interval?"

No, if two events are simultaneous in any frame, the interval between them is a space-like one, not a time-like one. If you're not familiar with the idea of the three categories of intervals, time-like, space-like and light-like, see the "Spacetime intervals in flat space" section of the wikipedia spacetime article. If you know the coordinates of two events in some inertial coordinate system, you can figure out the coordinate differences $\Delta x$, $\Delta y$, $\Delta z$ and $\Delta t$ between the events (i.e. if the first event had x-coordinate $x_1$ and the second had $x_2$, then $\Delta x = x_2 - x_1$), and then the spacetime interval can be defined as $s^2 = \Delta x^2 + \Delta y^2 \Delta z^2 - c^2\Delta t^2$. If $s^2$ is negative the interval is time-like (and the absolute value of $s$ is the proper time between the two points along the world line of an inertial object that passes through both, so this would be the actual time measured on this object's clock between two events which both occurred right next to the object), if $s^2$ is positive the interval is space-like (and $s$ is then the proper distance along a straight line path joining the two points--see my answer here for a little more on what "proper distance" means physically), and if $s^2$ is zero the interval is light-like (meaning only something moving at the speed of light could have both events on its world line).

Also, note that this interval is frame-invariant, which means the sense that if you switched to a different inertial coordinate systems, then even though the individual values of $\Delta x$, $\Delta y$, $\Delta z$ and $\Delta t$ could change, $s^2$ would be unchanged. This is analogous to the geometric case where if you describe two points in a 2D Euclidean plane using two Cartesian coordinate systems with differently-oriented $x$ and $y$ axes, the values of $\Delta x$ and $\Delta y$ may differ depending on the coordinate system, but the distance $d$ between the points, determined by the Pythagorean theorem to be $d^2 = \Delta x^2 + \Delta y^2$, would be the same in both coordinate systems.

Since the interval is frame-invariant, then if you calculate two events to have a space-like separation in the coordinates of one inertial frame, they will have the same space-like separation in all inertial frames, and thus it's meaningless to ask what the proper time between these two events is. And if the events are simultaneous in at least one frame, then in that frame $\Delta t = 0$, so naturally $s^2$ will be positive and the interval must be space-like.

"Can we take the time interval between the two events 1 and 2 as observed by the mentioned observer as proper time interval?"

No, if two events are simultaneous in any frame, the interval between them is a space-like one, not a time-like one. If you're not familiar with the idea of the three categories of intervals, time-like, space-like and light-like, see the "Spacetime intervals in flat space" section of the wikipedia spacetime article. If you know the coordinates of two events in some inertial coordinate system, you can figure out the coordinate differences $\Delta x$, $\Delta y$, $\Delta z$ and $\Delta t$ between the events (i.e. if the first event had x-coordinate $x_1$ and the second had $x_2$, then $\Delta x = x_2 - x_1$), and then the spacetime interval can be defined as $s^2 = \Delta x^2 + \Delta y^2 \Delta z^2 - c^2\Delta t^2$. If $s^2$ is negative the interval is time-like (and the absolute value of $s$ is the proper time between the two events along the world line of an inertial object that passes through both, so this would be the actual time measured on this object's clock between two events which both occurred right next to the object), if $s^2$ is positive the interval is space-like (and $s$ is then the proper distance along a straight line path joining the two events--see my answer here for a little more on what "proper distance" means physically), and if $s^2$ is zero the interval is light-like (meaning only something moving at the speed of light could have both events on its world line).

Also, note that this interval is frame-invariant, which means that if you switched to a different inertial coordinate systems, then even though the individual values of $\Delta x$, $\Delta y$, $\Delta z$ and $\Delta t$ could change, $s^2$ would be unchanged. This is analogous to the geometric case where if you describe two points in a 2D Euclidean plane using two Cartesian coordinate systems with differently-oriented $x$ and $y$ axes, the values of $\Delta x$ and $\Delta y$ may differ depending on the coordinate system, but the distance $d$ between the points, determined by the Pythagorean theorem to be $d^2 = \Delta x^2 + \Delta y^2$, would be the same in both coordinate systems.

Since the interval is frame-invariant, then if you calculate two events to have a space-like separation in the coordinates of one inertial frame, they will have the same space-like separation in all inertial frames, and thus it's meaningless to ask what the proper time between these two events is. And if the events are simultaneous in at least one frame, then in that frame $\Delta t = 0$, so naturally $s^2$ will be positive and the interval must be space-like.

"Can we take the time interval between the two events 1 and 2 as observed by the mentioned observer as proper time interval?"

No, if two events are simultaneous in any frame, the interval between them is a space-like one, not a time-like one. If you're not familiar with the idea of the three categories of intervals, time-like, space-like and light-like, see the "Spacetime intervals in flat space" section of the wikipedia spacetime article. If you know the coordinates of two events in some inertial coordinate system, you can figure out the coordinate differences $\Delta x$, $\Delta y$, $\Delta z$ and $\Delta t$ between the events (i.e. if the first event had x-coordinate $x_1$ and the second had $x_2$, then $\Delta x = x_2 - x_1$), and then the spacetime interval can be defined as $s^2 = \Delta x^2 + \Delta y^2 \Delta z^2 - c^2\Delta t^2$. If $s^2$ is negative the interval is time-like (and the absolute value of $s$ is the proper time between the two points along the world line of an inertial object that passes through both, so this would be the actual time measured on this object's clock between two events which both occurred right next to the object), if $s^2$ is positive the interval is space-like (and $s$ is then the proper distance along a straight line path joining the two points--see my answer here for a little more on what "proper distance" means physically), and if $s^2$ is zero the interval is light-like (meaning only something moving at the speed of light could have both events on its world line).

Also, note that this interval is frame-invariant, which means the sense that if you switched to a different inertial coordinate systems, then even though the individual values of $\Delta x$, $\Delta y$, $\Delta z$ and $\Delta t$ could change, $s^2$ would be unchanged. This is analogous to the geometric case where if you describe two points in a 2D Euclidean plane using two Cartesian coordinate systems with differently-oriented $x$ and $y$ axes, the values of $\Delta x$ and $\Delta y$ may differ depending on the coordinate system, but the distance $d$ between the points, determined by the Pythagorean theorem to be $d^2 = \Delta x^2 + \Delta y^2$, would be the same in both coordinate systems.

Since the interval is frame-invariant, then if you calculate two events to have a space-like separation in the coordinates of one inertial frame, they will have the same space-like separation in all inertial frames, and thus it's meaningless to ask what the proper time between these two events is.

"Can we take the time interval between the two events 1 and 2 as observed by the mentioned observer as proper time interval?"

No, if two events are simultaneous in any frame, the interval between them is a space-like one, not a time-like one. If you're not familiar with the idea of the three categories of intervals, time-like, space-like and light-like, see the "Spacetime intervals in flat space" section of the wikipedia spacetime article. If you know the coordinates of two events in some inertial coordinate system, you can figure out the coordinate differences $\Delta x$, $\Delta y$, $\Delta z$ and $\Delta t$ between the events (i.e. if the first event had x-coordinate $x_1$ and the second had $x_2$, then $\Delta x = x_2 - x_1$), and then the spacetime interval can be defined as $s^2 = \Delta x^2 + \Delta y^2 \Delta z^2 - c^2\Delta t^2$. If $s^2$ is negative the interval is time-like (and the absolute value of $s$ is the proper time between the two points along the world line of an inertial object that passes through both, so this would be the actual time measured on this object's clock between two events which both occurred right next to the object), if $s^2$ is positive the interval is space-like (and $s$ is then the proper distance along a straight line path joining the two points--see my answer here for a little more on what "proper distance" means physically), and if $s^2$ is zero the interval is light-like (meaning only something moving at the speed of light could have both events on its world line).

Also, note that this interval is frame-invariant, which means the sense that if you switched to a different inertial coordinate systems, then even though the individual values of $\Delta x$, $\Delta y$, $\Delta z$ and $\Delta t$ could change, $s^2$ would be unchanged. This is analogous to the geometric case where if you describe two points in a 2D Euclidean plane using two Cartesian coordinate systems with differently-oriented $x$ and $y$ axes, the values of $\Delta x$ and $\Delta y$ may differ depending on the coordinate system, but the distance $d$ between the points, determined by the Pythagorean theorem to be $d^2 = \Delta x^2 + \Delta y^2$, would be the same in both coordinate systems.

Since the interval is frame-invariant, then if you calculate two events to have a space-like separation in the coordinates of one inertial frame, they will have the same space-like separation in all inertial frames, and thus it's meaningless to ask what the proper time between these two events is. And if the events are simultaneous in at least one frame, then in that frame $\Delta t = 0$, so naturally $s^2$ will be positive and the interval must be space-like.